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## Karnataka 2nd PUC Physics Question Bank Chapter 3 Current Electricity

### 2nd PUC Physics Current Electricity One Marks Questions and Answers

Question 1.

What constitutes an electric current?

Answer:

The flow of electrons carrying electric charge constitutes an electric current.

Question 2.

What is the net flow of charge at the given cross section of a conductor?

Answer:

Zero.

Question 3.

Define electric current.

Answer:

The rate of flow of charge through any cross-section of a conductor is defined as electric current.

Question 4.

Mention the S.I. unit of current.

Answer:

ampere (A).

Question 5.

What is the direction of conventional current?

Answer:

The direction in which electric field is applied determines the direction of flow of positive charge. This direction is known as conventional current.

Question 6.

What is the direction of flow of electrons in a conductor?

Answer:

The direction of flow of electrons in a conductor is opposite to the direction of conventional current.

Question 7.

Define drift velocity.

Answer:

Drift velocity is defined as the average velocity with which electrons drift in a direction opposite to the direction of the applied electric field.

Question 8.

Define relaxation time or mean free time.

Answer:

The small time interval between two successive collisions between electrons and ions in a lattice is known as relaxation time.

Question 9.

Define electron mobility. (March 2014)

Answer:

Electron mobility is defined as the ratio of drift velocity of electron to the applied electric field.

Question 10.

What is the order of relaxation time?

Answer:

The order of relaxation time is 10^{-14} s.

Question 11.

Mention the SIU of electron mobility.

Answer:

The SIU of electron mobility is m^{2}V^{-1}s^{-1}.

Question 12.

How does drift velocity depend on electric current?

Answer:

Drift velocity of electrons is directly proportional to electric current.

Question 13.

What is the average electron thermal velocity at room temperature?

Answer:

The average electron thermal velocity at room temperature is 10^{6} ms^{-1}.

Question 14.

What is the order of drift velocity of electrons?

Answer:

The drift velocity of electrons is of the order of 10^{-3} ms^{-1}.

Question 15.

State Ohm’s law.

Answer:

Ohm’s law: When all the physical conditions surrounding the conductor remain unaltered, the electric current is directly proportional to the potential difference between its ends.

Question 16.

Define electrical resistance.

Answer:

Electrical resistance is defined as the ratio of the potential difference between the ends of the conductor to the rate of flow of charge.

Question 17.

Mention the SIU of electrical resistance.

Answer:

ohm (Ω).

Question 18.

Define the unit one ohm.

Answer:

If one volt of potential difference is maintained between the ends of a conductor and if one ampere of current results in the conductor, then the electrical resistance offered by the conductor is said to be one ohm.

Question 19.

How does the resistance of a conductor depend on its length?

Answer:

The resistance of a conductor is directly proportional to the length of the conductor. R α l

Question 20.

How does electrical resistance depend on the area of cross-section of a conductor?

Answer:

Electrical resistance is inversely proportional to the area of cross-section of a conductor.

Note : \(R \propto \frac{1}{A} \propto \frac{1}{d^{2}} \propto \frac{1}{r^{2}}\) where ‘r’ is the radius of cross section of cylindrical conductor.

Question 21.

Define electrical conductance?

Answer:

The reciprocal of electrical resistance is called as electrical conductance.

Question 22.

Mention the SI units of electrical conductance?

Answer:

siemen or mho.

Question 23.

What are ohmic devices?

Answer:

Ohmic devices are those devices which obey Ohm’s law.

Question 24.

Give an example for a ohmic device.

Answer:

Conductor at room temperature.

Question 25.

A graph of electric current along the y-axis and potential difference along the x-axis gives a straight line passing through the origin. What does the slope determine?

Answer:

Electrical resistance (R).

Question 26.

What are non ohmic devices? ‘

Answer:

Devices or circuit elements which do not obey Ohm’s law are called non ohmic devices.

Question 27.

Write the dimensional formula for electrical resistance.’

Answer:

[R] = [ML^{2}T^{-3}A^{-2}]

Question 28.

Define resistivity.

Answer:

Resistivity of the material of a conductor may be defined as numerically equal to the resistance of unit area of section and unit length of that conductor.

Question 29.

Mention the SIU of resistivity.

Answer:

The SIU of resistivity is ohm-meter (Ω m).

Question 30.

Mention the dimensional formula for resistivity.

Answer:

[p] = [MZ^{3}T^{-3}A^{-2}]

Question 31.

What is meant by electrical conductivity?

Answer:

The reciprocal of electrical resistivity of the material of a conductor is known as electrical conductivity.

Question 32.

Mention the SI unit of electrical conductivity.

Answer:

(Ω m)^{-1} or mho-m^{-1}.

Question 33.

Write the equivalent mathematical form for Ohm’s law.

Answer:

\(\vec{j}=\sigma \overrightarrow{\mathrm{E}}\), where, j – current density, E – electric field intensity and σ = conductivity.

Question 34.

Express the range of resistivity of conductors.

Answer:

The range of resistivity varies from 10^{-8} Ωm to 10^{-6} Ωm for conductors.

Question 35.

Express the range of resistivity of semiconductors.

Answer:

The range of resistivity varies from 10^{-5} Ωm to 10^{3} Ωm for semiconductors.

Question 36.

Mention the range of resistivity of insulators.

Answer:

The range of resistivity varies from 10^{5}Ωmto 10^{16}Ωm (water & fused quartz) for insulators.

Question 37.

Draw a graph of resistivity of copper as a function of absolute temperature.

Answer:

Question 38.

Draw a graph of resistivity of nichrome as a function of absolute temperature.

Answer:

Question 39.

Plot a graph of resistivity of a semiconductor as a function of absolute temperature.

Answer:

Question 40.

The first three colour bands are Red, Red, Orange. What will be the value of resistance of the resistor?

Answer:

22 × 10^{3}Ω±20%.

Question 41.

Define temperature coefficient of resistance.

Answer:

Temperature coefficient of resistance may be defined as the ratio of change in the resistance per degree change in temperature of a conductor, to its resistance at 0°C.

Question 42.

What is meant by ‘Critical temperature?

Answer:

Critical temperature is that lower temperature at which a Conductor shows the property of super conductivity.

Question 43.

What are super conductors?

Answer:

Conductors whose electrical resistance becomes zero, near zero K are called super conductors.

Question 44.

A carbon resistor is marked Red, Red, Red, Gold. What is the value of its resistance?

Answer:

R = 22 × 10^{2}Ω ± 5% i.e., R = 22KΩ ± 0.11KΩ.

Question 45.

Compare R_{s} with R_{p} for ‘n’ identical resistors connected in series and parallel combinations.

Answer:

R_{s} = n^{2}R_{p}.

Question 46.

Define EMF of a cell.

Answer:

Electromotive force of a cell may be defined as its terminal potential difference, when no electric current is drawn from it.

Question 47.

What amount of energy is derived from a cell by a capacitor, to charge it completely?

Answer:

The energy drawn from the cell is 50% of the available energy from the cell.

Question 48.

Define internal resistance of a cell.

Answer:

Internal resistance of a cell may be defined as the opposition offered by the electrolyte to the flow of charges.

Question 49.

Define one volt of emf

Answer:

One volt of emf may be defined as the amount of energy equal to one joule required to drive one coulomb of charge once around the closed circuit.

Question 50.

What happens to the effective resistance when two resistors are connected in series?

Answer:

The effective resistance of two or a number of resistors in series combination increases and is greater than the maximum value of the individual resistances used.

Question 51.

Two resistors are connected in parallel. What will happen to its equivalent resistance?

Answer:

The effective resistance becomes minimum and lower than the least value of resistance of the resistors.

Question 52.

Define equivalent resistance of a number of resistors connected in a series or parallel combination.

Answer:

Equivalent resistance of a number of resistors connected in a series or parallel may be defined as the single resistance which draws the same amount of current as that of the combination of resistors.

Question 53.

Give the expression for equivalent resistance of two resistors connected in parallel.

Answer:

R_{p} = \(\frac{R_{1} R_{2}}{R_{1}+R_{2}}\) where R_{1}, & R_{2} are the individual resistances.

Question 54.

What is the reciprocal of resistance called?

Answer:

Conductance. The unit of conductance is expressed as ohm^{-1} or mho or siemen (S).

Question 55.

What is the reciprocal of resistivity called?

Answer:

Conductivity expressed as siemen m^{-1} or Sm^{-1}.

Question 56.

Why is that the terminal potential difference is always less than the e.m.f. of a cell?

Answer:

Since V = E – Ir, a part of energy is spent to over come the internal resistance offered by the electrolyte of the cell.

Question 57.

What is the unit of e.m.f?

Answer:

volt(V)

Question 58.

Give the colour code for 1 kΩ ± 10% tolerance resistor.

Answer:

Brown, Black, Red, Silver

Question 59.

In the following graph, say whether T_{1} > T_{2} or not.

Answer:

Since slope of (1) > slope of (2),

K_{1} > K_{2}, R_{1} < R_{2}, since R temperature. T_{1} < T_{2}

Question 60.

How does resistivity depend on number density of electrons?

Answer:

p α \(\frac{1}{n}\).

Question 61.

The length of a conductor is increased two folds. What will happen to the resistivity of the material?

Answer:

Resistivity will remain the same.

Question 62.

Why are Manganin and Constantan used in making resistance coils?

Answer:

The temperature coefficients of resistance of Manganin and Constantan are very small and their resistivity very high. Therefore these materials are used in making resistance coils.

Question 63.

Define the term current density.

Answer:

Current density is defined as the ratio of electric current to the area of cross section of the conductor.

Question 64.

What is an electrical network?

Answer:

An electrical network is a combination of various circuit elements and sources of emf connected together.

Question 65.

What is a node?

Answer:

A node or a junction is a point in a network at which several conductors are connected or joined.

Question 66.

What is a mesh or loop?

Answer:

A mesh is a closed path within the network in which different elements of a mesh may or may not carry different currents.

Question 67.

What principle is involved in Kirchhoff’s first law?

Answer:

Kirchhoff’s first law is based on the law of conservation of charge.

Question 68.

What is the significance of Kirchhoff’s voltage law?

Answer:

Kirchhoff’s voltage law is based on the law of conservation of energy.

Question 69.

Can Kirchhoff’s laws be applied to AC circuits?

Answer:

Yes.

Question 70.

What is the principle of a metre bridge?

Answer:

A metre bridge is based on the balanced condition of Wheatstone’s network, so that the unknown resistance can be determined.

Question 71.

Write the balanced condition of Wheatstone’s network. (March 2014)

Answer:

For a balanced condition i_{g} = 0 such that \(\frac{P}{Q}=\frac{R}{S}\) where resistors of resistance P, Q, S, R are connected in cyclic order.

Question 72.

Why are metal strips thick in a metre bridge?

Answer:

Metal strips are generally thick because many electrical connections can be derived without much current loss.

Question 73.

What is generally connected in the right gap of a metre bridge?

Answer:

The right gap of a metre bridge is generally connected to a standard resistance box.

Question 74.

If the galvanometer in the Wheatstone’s network is removed under the balanced condition, then say whether the network is still balanced or not.

Answer:

The network will be still balanced.

Question 75.

In a balanced Wheatstone’s network, galvanometer and cell are interchanged. Will the network be still balanced?

Answer:

Yes.

Question 76.

In a balanced Wheatstone’s network, the galvanometer resistance is increased by 20 Ω. What happens to the balance of the network?

Answer:

The network will remain balanced.

Question 77.

At what point of a closed circuit, conservation of charge is valid?

Answer:

At the junction or node.

Question 78.

Name the instrument which can measure the emf of a cell and its internal resistance accurately.

Answer:

Potentiometer.

Question 79.

Name the physical quantity which gives the rate of flow of charge per unit area normal to the flow. Is current density a scalar or a vector?

Answer:

Current density. (\(\left(\vec{j}=n q \vec{v}_{d}\right)\)). The current density is a vector quantity.

Question 80.

Name any one device that works based on balanced condition of the Wheatstone’s bridge or network.

Answer:

Metre bridge.

Question 81.

What will be the resistance of a semiconductor near absolute zero.

Answer:

Infinity.

Question 82.

What will be the resistance of a conductor near absolute zero?

Answer:

Zero.

Question 83.

A resistor draws energy from a battery. Where does this energy come from?

Answer:

Energy supplied to the load resistor comes from chemical energy of the electrolyte.

Question 84.

What are semiconductors?

Answer:

Substances whose electrical conductivity at high temperatures behaves as conductors and as insulators at low temperatures, are called semiconductors.

Question 85.

What is meant by electrical grounding or earthing?

Answer:

When a charged conductor or body comes in contact with the ground, charges (electrons) flow from the ground to the positive body and to the ground from the negative body.

Question 86.

What is Super conductivity?

Answer:

The phenomenon of a conductor losing the electrical resistance at a temperature called critical temperature, is known as superconductivity.

### 2nd PUC Physics Current Electricity Two Marks Questions and Answers

Question 1.

Mention any two effects of current.

Answer:

Effects of current are,

- Magnetic effect,
- Mechanical effect and
- Chemical effect.

Question 2.

Mention any two factors on which the internal resistance of the cell depends.

Answer:

The internal resistance of a cell depends on

- the nature of the electrolyte
- nature of electrodes
- temperature.

Question 3.

Mention the unit of current density. Say whether current density ¡s a scalar or a vector physical quantity.

Answer:

The SI unit of current density is Am^{-2}. Current density is a vector physical quantity.

Question 4.

Write the dimensional formula for (i) voltage (ii) resistivity.

Answer:

Dimensional formula for,

Question 5.

State Kirchhoff’s laws of Electrical network.

Answer:

Kirchhoff’s current law: The algebraic sum of currents entering and leaving a node of an electrical network is zero.

Kirchhoff’s voltage law: The algebraic sum of ‘IR’ products is equal to the algebraic sum of emfs, in any closed mesh of an electrical network.

Question 6.

P, Q, S, R are four resistances arranged in a cyclic order in a Wheatstone’s network. Under a balanced condition, what will be the potential difference between the junction of P and Q and R and S?

Answer:

For a balanced condition, i_{g} = 0

Hence voltage between the diagonal junctions will be zero.

Question 7.

What are electrical conductors? Give an example.

Answer:

Substances which readily allow electricity to pass through them are called conductors, e.g.: Metals, human and animal bodies, earth, moist air are conductors.

Question 8.

What are insulators? Give an example.

Answer:

Substances which oppose the passage of electricity through them are called insulators, e.g.: Glass, Porcelain; Plastic, Ceramic, Nylon, Wood, Paper.

Question 9.

Two bulbs 110 V / 10 W and 110 V / 40W are connected in series and across a 110 V. Which of these will glow brighter? Explain why.

Answer:

Resistance of lower wattage > Resistance of higher wattage for a given voltage. Since V_{1} : V_{2} :: R_{1} : R_{2}, the lower wattage glows brighter.

Question 10.

Two heaters rated 220 V/1 kW and 220 V / 3 kW are connected in parallel across a 220V source. Which of these will heat up faster? Explain why.

Answer:

The heater rated 220V/3 kW heats faster than the heater rated 220 V /1 kW because branch current across a higher wattage is more than that across a lower wattage.

Question 11.

If ε_{1}, ε_{2}, ε_{3} …… ε_{n} represent emfs of a number of cells of internal resistances r_{1}, r_{2}, …… r_{n} then mention their equivalent emf and internal resistance when these are connected in series.

Answer:

Question 12.

If a number of cells are connected in parallel then write their equivalent emf and internal resistance.

Answer:

Question 13.

In the network shown below, find I.

Answer:

At junction A, (3 + 2 – 1) = 4A. At junction B, (4 + 1) = 5 A. Current through BC is 5A.

∴ I = (5 – 4) = 1 A.

Question 14.

What is a potentiometer?

Answer:

A potentiometer is a long wire of uniform thickness across which a standard cell is connected and is used as a device to measure emfs and internal resistance of the given cells.

Question 15.

Give the formula to determine internal resistance of a cell using a potentiometer.

Answer:

r = R\(\left(\frac{l_{1}}{l_{2}}-1\right)\) where l_{1} – balancing length with key connected to external resistance R in parallel to the cell is open.

l_{2} – balancing length with key connected to external resistance R is closed.

Question 16.

How does the emf of a cell depend on the balancing length obtained in a potentiometer? Give the expression for ratio of emfs in terms of balancing lengths.

Answer:

Question 17.

Give the expression for power wasted in connecting wires.

Answer:

P_{c} = I^{2}R_{c} i.e., P_{c} = \(\frac{\mathrm{P}^{2} \mathrm{R}_{\mathrm{c}}}{\mathrm{V}^{2}}\). Where, P is power delivered to the load, P_{c} = power wasted as heat, R_{c} – resistance of connecting wires.

Question 18.

Give any two practical limitations of Ohm’s law. (March 2015)

Answer:

- Ohm’s law cannot be applied for non-ohmic devices such as semiconductors, electrolytes of a cell, discharge tubes, thermionic valve devices.
- Ohm’s law cannot be applied for conductors for varying temperatures, for temperatures near absolute temperature and for higher temperatures.

### 2nd PUC Physics Current Electricity Three Marks Questions and Answers

Question 1.

A uniform wire of resistance 9Ω is bent to form an equilateral triangle. What is the effective resistance between any two corners of the triangle.

Answer:

Question 2.

State and explain Ohm’s law.

Answer:

Ohm’s law.: When all the physical conditions surrounding the conductor remain unaltered, the electric current in a conductor is directly proportional to the potential difference between its ends.

If ‘F is the current, ‘ V’, the p.d. between the ends of the conductor then

I α V

i.e., I = KV where ‘K’ is known as electrical conductance.

V = IR

where, R = \(\frac{1}{K}\) is known as the electrical resistance of the conductor.

Question 3.

Define temperature coefficient of resistance of a conductor. Give the expression for temperature coefficient.

Answer:

The temperature coefficient of resistance of a material of conductor may be defined as the ratio of change in its resistance per kelvin raise of temperature to its resistance at 0°C

Where, R_{0} – Resistance of the conductor at 0°C

for two differents resistences at l_{1} and l_{2} temperature.

Question 4.

Derive the expression for current in a simple circuit based on Ohm’s law.

Answer:

From the circuit, E = IR + Ir

i.e., I = \(\frac{E}{R+r}\)

Where ‘r’ is the internal resistance of a cell.

Question 5.

If 10 Ω is unplugged from a standard resistance box and 33\(\frac { 1 }{ 3 }\) cm is the balancing length obtained in the metre bridge, then calculate the resistance of the wire.

Answer:

Question 6.

Four resistors of resistances 2Ω, 4Ω, 6Ω and 2Ω are connected in a cyclic order of a Wheatstone’s network. What resistance should be connected with 2Ω in the arm AD so as to balance the network?

Answer:

For a balanced network,

Since x > 2Ω, a resistor of ‘ 1Ω’ resistance is connected in series with 2Ω in the arm AD to balance the network.

Question 7.

Four resistors 4Ω, 8Ω, 18Ω and 6Ω are connected in a cyclic order to form a Wheatstone’s network. How and what value of resistance should be connected in the 18 Ω branch to. balance the network?

Answer:

For the balanced condition, \(\frac{P}{Q}=\frac{R}{S}\)

i.e., \(\frac{4}{8}=\frac{6}{x^{8}}\)

∴ x = \(\frac{8 \times 6}{4}\) = 12Ω 12Ω < 18Ω

A resistor y is to connected in parallel to 18 Ω to make it 12 Ω

= \(\frac{18 y}{18+y}\) = 12

216 + 12y = 18y

6y = 216

∴ y = 36 Ω

A resistor of resistance 36 Ω should be connected in parallel across 18 Ω to balance the network.

Question 8.

Write a note on Wheatstone’s network.

Answer:

Wheatstone’s network comprises of four resistors connected to form the four arms of a square. A galvanometer is connected across one pair of opposite diagonal corners and the other opposite diagonal corners are connected to the emf source.

For a balanced condition, P and Q are adjacent resistances and R and S form another pair of resistances. If three of the four resistances are known, then the resistance of the fourth arm can be calculated.

Question 9.

If two parallel resistors P and R of Wheatstone’s network, are short circuited, then find the equivalent resistance of the circuit, when G = S = Q = 10 Ω.

Answer:

From the connection, it is clear that Q || G || S.

Since G = S = Q = 10 Ω,

Question 10.

Show that j = σE where symbols have their usual meanings.

Answer:

We know from the Ohm’s law, I = KV

where, I – electric current, K – conductance and V- potential difference

Dividing I = KV by ‘A’ on both sides we get,

\(\frac{I}{A}=\frac{\mathrm{K}}{A} \cdot V\)

bt definition electric current density j = \(\frac{I}{A}\)

Hence j = \(\left(\frac{\mathrm{K}}{A}\right) V\). But electric field,

E = \(\frac{V}{l}\) or V = El

Hence j = \(\left(\frac{\mathrm{K}}{A}\right)\)(El) where σ = \(\frac{\mathrm{K} l}{A}\) is known as; the electrical conductivity.

Hence j = σE. In a vector form, the relation is written as \(\vec{j}=\sigma \vec{E}\)

Question 11.

Distinguish between Current and Current density.

Answer:

### 2nd PUC Physics Current Electricity Five Marks Questions and Answers

Question 1.

Derive an expression for electrical conductivity.

Answer:

Let ‘n’ be the number density of electrons. Let ‘L’ be the length of the conductor, ‘A’ be the area of cross-section of the conductor.

Let v_{d} be the drift velocity of the electrons. Let ∆x be a small length.

Electrons drift in a direction opposite to the electric field.

Number of electrons in ‘∆x’ and area of cross-section ‘A’ is equal to (A∆x)n.

Charge on these electrons = (nA∆x)e

If ∆t is the time taken for effective displacement of electrons then rate of flow charge = nAe\(\left(\frac{\Delta x}{\Delta t}\right)\) By definition, electric current I = rate of flow of charge.

i.e., I = nAev_{d} …… (1)

where, average velocity of electrons with which it drifts against the direction of electric field is known as drift velocity (v_{d}).

Let ‘a’ be the acceleration of electrons. Let ‘E’ be the electric field intensity. Force on electrons,

F = ma

i.e., eE = ma

but = a = v_{d}x where ‘τ’ is the relaxation time. Relaxation time represents the average time taken for two successive collisions of electrons and ions in the lattice.

We also know that electric potential difference between the ends of the conductor V = EL

Question 2.

State the laws of resistance of a conductor. Define specific resistance of material of a conductor. Mention the SI unit of specific resistance.

Answer:

Resistance ‘R’of a conductor is,

(i) directly proportional to the length of the conductor.

(ii) inversely proportional to the area of cross-section of the conductor.

where, ‘ p’ is the resistivity or specific resistance of the conductor.

For one metre cube of the material, L = 1 m, A = lm^{2} hence R = p.

Specific resistance of material of the conductor may be defined as numerically equal to the resistance of the material taken in the form of a cube of side one metre.

The SIU of specificresistance is ‘Ωm’. Specific resistance is different for different materials and hence it is an unique physical quantity.

Question 3.

Write a note on the variation of resistance of a metallic conductor with temperature.

Answer:

The resistance of a conductor increases or decreases with the increase or decrease of temperature.

R = R_{0}(1 + αt + βt^{2} + γt^{3} + ) where α, β, γ are called the I order, II order and III order

temperature coefficients respectively. However, β, γ, ……… are far less than‘ α’.

Hence R = R_{0}(1 + αt), to a near degree of approximation. Electrical resistance varies linearly with the resistance due to collision between electrons and oscillating ions in the lattice.

We can show that α = \(\frac{R-R_{0}}{R_{0} t} K^{-1}\) Temperature coefficient resistance of a conductor is

defined as the ratio of change in the resistance per degree kelvin change in the temperature to its resistance at 0°C or 273 K.

A graph of resistance along they-axis and temperature along the x-axis, gives a straight line with ay-intercept ‘R_{0}’and slope of the line‘R_{0}α

α = \(\left(\frac{1}{R_{0}}\right)\) (slope of the line).

The x-intercept on the ‘-x’ axis indicates that the resistance of the conductor vanishes at a temperature called transition or critical temperature. Conductors become super conductors at the critical temperature and below it. The phenomenon of conductors losing all the resistance to the flow of charges at the critical temperature and below it is known as super conductivity.

Question 4.

Obtain an (expression for equivalent resistance of two resistors connected in a series combination.

Answer:

Resistors are said to be connected in series, when

(i) the same current passes through each of the resistors

(ii) net voltage gets divided such that V = V_{1} + V_{2} From Ohm’s law V= IR.

ie.,V_{1} = IR_{1} ; V_{2} = lR_{2}

∴ V = I(R_{1} + R_{2})

or R_{1} = \(\left(\frac{V}{I}\right)\) = R_{1} + R_{2}

For two resistor sin series R_{1} = R_{1} + R_{2}

Hence when a number of resistors are connected in series, their equivalent resistance is equal to the algebraic sum of the individual resistances.

Note: For n number of resistors in series R_{s} = \(\sum_{i=1}^{n}\)R_{1}

For n identical resistors in series R_{s} = nR.

Question 5.

Obtain an expression for equivalent resistance of two resistors connected in parallel. (March 2014)

Answer:

Resistors are said to be connected in parallel when,

(i) Voltage remains common

(ii) Current branches out in each of the resistors such that

For two resistors in parallel \(\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

When resistors are connected in parallel, the reciprocal of their equivalent resistance is ‘ equal to the algebraic sum of reciprocals of individual resistances.

Note: For V resistors in parallel, \(\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots \ldots+\left(\frac{1}{R_{n}}\right)\)

For V identical resistors in parallel R_{p} = \(\frac{R}{n}\)

Question 6.

State and explain Kirchhoff’s laws of electrical network.

Answer:

Kirchhoff’s current law: The algebraic sum of currents entering and leaving a junction in an electrical network is zero.

i.e., ΣI = 0

Kirchhoff’s current law works on the principle of law of conservation of charge. Currents entering the node are taken as +ve and leaving the node are taken as

In the circuit given above, +I_{1}, + I_{2} – I_{3} – I_{4} = 0

i.e., I_{1} + I_{2} = I_{3} + I_{4}

Hence algebraic sum of currents entering the node is equal to the algebraic sum of currents leaving the node.

Kirchhoff’s voltage law: The algebraic sum of ‘IR’ products is equal to the algebraic sum of emfs present in a mesh of an electrical network.

ΣIR – ΣE

This law is based on the law of conservation of energy. ‘IR’ products are taken +ve if the chosen direction is the same as current in the resistor. ‘E’ is taken as +ve when the chosen direction leaves the+ve terminal of the emf source.

Question 7.

Arrive at the balanced condition of a Wheatstone’s network.

Answer:

Applying KCL to junction B and D

I_{1} = I_{3} + I_{g} ………. (1)

I_{4} = I_{2} + I_{g} ………. (2)

Applying KVL to the mesh ABDA,

I_{1}P + I_{g}G – I_{2}R = 0 ……… (3)

i.e., I_{1}P + I_{g}G = I_{2}R ……… (4)

Applying KVL to the mesh BCDB, I_{3}Q – I_{4}S – I_{g}G = 0

For a balanced condition of Wheatstone’s network, I_{g} = 0.

(1) and (2) may be written as I_{1} = I_{3} and I_{3} = I_{4}

(3) and (4) may be written as

I_{1}P = I_{2}R ….. (5)

and I_{3}Q = I_{4}S ….. (6)

Question 8.

Obtain an expression for the equivalent emf of two cells connected in series.

Answer:

Let E_{1}, E_{2} represent emfs of cells having internal resistances r_{1}, r_{2} respectively.

Let V be the terminal p.d across the load resistance R

∴ for a series combination of a number of cel Is,

We can easily note that for a series combination of cells, equivalent emf is the algebraic sum of individual emfs, and equivalent internal resistance of series combination is the algebraic sum of their internal resistances.

Note : For a number of identical cells in series combination,

Question 9.

Obtain an expression for the equivalent emf and internal resistance of two cells connected in parallel.

Answer:

Let I_{1}, I_{2} represent branch currents.

Let V be the common potential, so that,

Comparing this with the terminal potential difference,

(ii) For number of identical cells

E – emf of each cell

r – internal resistance of each cell.

### 2nd PUC Physics Current Electricity List of Formulae

### 2nd PUC Physics Current Electricity Numericals and solutions

Question 1.

In the given circuit, calculate the (i) effective resistance between A and B (ii) current through the circuit and (iii) current through 3 resistor.

Answer:

Equivalent circuit may be written as

Question 2.

A battery of internal resistance 3 Ω is connected to 20 Ω resistor and the potential difference across the resistor is 10V. If another resistor 30Ω is connected in series with the first resistor and battery is again connected to the combination, then calculate the e.m.f and terminal potential difference across the combination. (March 2014)

Answer:

Case (i):

Terminal potential difference V = \(\frac{E R}{R+r}\)

i.e., 10 = \(\frac{E(20)}{20+3}\)

∴ E = 11.5V.

Case (ii):

Question 3.

A network of resistors is connected to a 12V battery as shown in the figure.

(a) Calculate the equivalent resistance of the network.

(b) Obtain current in 12 Ω and 6 Ω resistors.

Answer:

Question 4.

Two cells of emf 2V and 4V and internal resistance 1Ω and 2Ω respectively arc connected in parallel so as to send the current in the same direction through an external resistance of 10Ω. Find the potential difference across 10Ω resistor.

Answer:

Given E_{1} = 2V, r_{1} = 1 Ω

E_{2} = 4V, r_{2} = 2 Ω

Question 5.

When two resistors are connected in series with a cell of cml 2V and negligible internal resistance, a current of (215)A flows in the circuit. When the resistances are in parallel, the main current ¡s (5/3)A. Calculate the resistances.

Answer:

Given E = 2 V, r = 0

For a series combination,

R_{1} = R_{1} + R_{2}, I = \(\frac{E}{R_{s}+r}\)

i.e., I = \(\frac{2}{R_{s}}=\frac{2}{5}\)

or R_{3} = 5Ω

or R_{1} + R_{2} = 5

For a parallel combination,

Solving R_{1} + R_{2} = 5 we get 2R = 6

and R_{1} – R_{2} = 1 or R_{1} = 3Ω

and R_{1} = 2Ω

Question 6.

Two resistors of resistances 2Ω and 4Ω are connected in parallel. Two more resistors 3Ω and 6Ω are also connected in parallel. These two combinations are in series with a battery of emf 5 V and internal resistance 0.7Ω Calculate current through the 6Ω resistor.

Answer:

Question 7.

Three equal resistors connected in series across a source of emf of negligible internal resistance together dissipate 10 W of power. What would be the power dissipated if the resistors are connected in parallel across the same source of e.m.f.

Answer:

Given three resistors are connected in series

i.e. R_{s} = 3R

Also given, power dissipation P = 10 W

i.e., 10 = \(\frac{V^{2}}{R_{s}}\) here V = E ; r = 0

i.e., V^{2} = 10 × 3R

For parallel combination R_{p} = \(\left(\frac{R}{3}\right)\) so that

\(P^{\prime}=\frac{V^{2}}{R_{p}}=\frac{3 V^{2}}{R}\)

i.e., P’ = 3(10) = 30 W

Question 8.

A copper wire has 3 × 10^{22} free electrons in 0.021 m length. The drift velocity of electrons is found to be 2 × 10^{-5} ms^{-1}, (i) How large a current will flow through the wire? (ii) How many electrons would pass through a given cross-section of the wire in one second?

Answer:

Since, I = ndev_{d}

We, get I = Nev_{d}

i.e., I = (1.429 × 10^{24})(1.6 × 10^{-19})(2 × 10^{-5})A

i.e., I = 4.5728 A.

(ii) Number of electrons crossing the conductor

Question 9.

A cube of side 0.05 m of a material of conductor is drawn into a wire of thickness 0.20 mm. If the specific resistance of the wire is 10^{-6} Ω m then calculate the resistance of the wire.

Answer:

Question 10.

If the number density of electrons in a conductor is 3 × 10^{20} and thickness of the conductor is 5 mm, then calculate the current in that conductor for a drift velocity of 2 × 10^{-4} ms^{-1}. If the voltage applied is 500 V, then calculate relaxation time. Given specific charge of electron = 1.76 × 10^{11} C kg^{-1}, length of the conductor = 10 m.

Answer:

Given : n = 3 × 10^{28} electrons / m^{3}, d = 5 × 10^{-3} m,

v_{d} = 2 × 10^{-4} ms^{-1}, V = 500 V, \(\frac{e}{m}\) = 1.76 × 10^{11} Ckg^{-1}

τ = ? L = 10 m

Question 11.

Two resistors 10 Ω and 25 Ω are connected in series and the combination is connected across a 25 V source of negligible internal resistance. A voltmeter of resistance 25 Ω is connected across the 10 Ω resistor. Find the reading in the voltmeter. If another voltmeter of a very large resistance replaces the voltmeter, then find title reading of voltmeter.

Answer:

p.d. across the 25 12 resistor = 0.778 × 25 = 19.44 V.

Voltmeter reading = 25 – 19.44 5.56 V.

In the absence of the 25 Ω voltmeter,

I = \(\frac{25}{10+25}\) = 0.7143 A

p.d. across the 25 Ω resistor = 0.7143 × 25 = 17.86 V.

New volt meter reading will be 25 – 17.86 = 7.14 V.

The difference in the readings of the voltmeters is

7.14 – 5.56 = 1.58 V

Question 12.

Two cells are connected in parallel such that maximum current is derived by the load resistor of 10Ω. If the sources of e.m.f. are 6V, 2Ω and 4V, 3Ω then calculate the power dissipation in the load resistor.

Answer:

Applying KCL to the node‘A’,

we get, I = I_{1} + I_{2} ….. (1)

Applying KVL to the mesh comprising 6V and 4V source

we get, 2I_{1} – 3I_{2} = 6 – 4

i.e., 2I_{1} – 3I_{2} = 2 …… (2)

Applying KVL to the mesh comprising 4V and 10 Q we get

3I_{2} + 10 I = 4

i.e., 3I_{2} + 10(I_{1} + I_{2}) = 4

10I_{1} + 13I_{2} = 4 ……. (3)

multiplying (2) by 5 and subtracting from (3) we get,

Hence, I_{1} = 0.6786 A

I_{2} = – 0.2143 A

Main current I = I_{1} + I_{2} = 0.4642 A ; P = I^{2}R = (0.4642)^{2} × (10) = 2.156 W

The power dissipation in 10Ω resistor is 2.156 W.

Question 13.

The positive terminals of two cells 3V, IΩ and 4V, 2Ω are connected to a wire of resistance 6Ω and their negative terminals to a wire of resistance 8Ω. The middle terminals of these wires are connected to a 10Ω resistor. Calculate the power ‘ dissipation in the 10Ω resistor.

Answer:

Applying KCL to the node A, I = I_{1} + I_{2}

Applying KVL to the mesh (1),

1(I_{1}) + 3I_{1} + 10(I_{1} + I_{2}) + 4I_{1} = 3

18I_{1} + 10I_{2} = 3 …..(1)

Applying KVL to the mesh (2),

2I_{2} + 3I_{2} + 4I_{2} + 10 (I_{1} + I_{2}) = 4

10I_{1} + 19I_{2} = 4

Multiplying equation (1) by 10 and equation (2) by 18 and subtracting the two we get,

Substituting for I_{2} in (1) we get

18I_{1} + 10(0.1735)= 3

18I_{1} = 3 – 1.735 = 1.265

∴ I_{1} = \(\frac{1.265}{18}\) = 0.0703A

Main current I = 0.1735 + 0.0703 = 0.2438 A.

Power dissipation in 10Q resistor = PR – (0.2438)^{2} × 10 = 0.594 W

Question 14.

Four resistors of resistances 2Ω, 1Ω, 3Ω, 4Ω are arranged in a cyclic order to form a Wheatstone’s network ABCD. The junctions B and D are connected to a galvanometer of resistance 2Ω. A current of 0.1 Centers the junction ‘A’. Calculate the current in the galvanometer.

Answer:

Applying KCL to the node A, I_{1} + I_{2} = 0.1

∴ I_{2} = 0.1 – I_{1} …. (1)

Applying KCL to the node B,

I_{1} = I_{3} + I_{g}

So that, I_{3} = I_{1} – I_{g} ….. (2)

Applying KCL’to the node D, I_{2} + I_{g} = I_{4}

substituting for I_{2}

I_{4} = 0.1 – I_{1} + I_{g}…… (3)

Applying KVL to the mesh ABDA,

2I_{1} + 2I_{g} – 4(0.1 – I_{1}) = 0

6I_{1} + 2I_{g} = 0.4 …. (4)

Applying KVL to the mesh BCDB we write

1I_{3} – 3(0.1 – I_{1} + I_{g}) – 2I_{g} = 0

+3I_{1} – 5I_{g} + (I_{1} – I_{g}) = 0.3

4I_{1} – 6I_{g} = 0.3

by using determinant method, equations (4) and (5) may be solved.

The current in the galvanometer is 4.545 mA from D to B.

Question 15.

ABCD forms the four arms of a Wheatstone’s network. The arms AB, BC, CD, DA are of 10Ω, 15Ω, 25Ω and 20Ω respectively. The junctions B and D are connected to the ends of galvanometer of resistance 25Ω. If the source voltage is 5V, r = 0, then calculate the current in the galvanometer. Assume the branch currents across AB, BC, CD, DA and BD as i_{1}, i_{3},- i_{4}, -i_{2}, and i_{g} respectively.

Answer:

Question 16.

A heater coil is rated 100W, 200V. It is cut into two identical parts. Both are connected together in parallel to the same source of 200V. Calculate the power dissipated per second in the new combination.

Answer:

Given : P = 100W, V = 200 V

Resistance of the heater coil = R = \(\frac{V^{2}}{P}\)

i.e., R = \(\frac{200 \times 200}{100}\) = 400 Ω

Resistance of each identical part of the coil R’ = \(\frac{R}{2}=\frac{400}{2}\) = 200Ω

Equivalent resistance of coils connected in parallel = \(\frac{R^{\prime}}{2}\) = 100Ω

Energy liberated when the combination is connected to the source of 200V supply

Question 17.

Twelve equal wires each of resistance V are joined to form a skeleton cube. The current enters at one corner and leaves at the diagonally opposite corner. Find the total resistance of between these corners.

Answer:

For a closed loop AKLMYXA,

2Ir + Ir + 2Ir = E

5Ir = E = I’R_{xy}

but, I’ = 61

∴ 5Ir = 6IR_{XY}

or R_{XY} = R_{eq} = \(\frac { 5 }{ 6 }\)r

Question 18.

Calculate the equivalent resistance of the following circuits (a) and (b).

Answer:

Question 19.

The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4Ω, then what is the maximum current that can be drawn from the battery?

Answer:

The current is maximum for an external load resistance R = 0.

Given, emf E = 12V, internal resistance r = 0.4Ω.

Question 20.

A battery of cmf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer:

Given, emf of the battery – 10 V(E), internal resistance = 3Ω(r)

We know that the current in the circuit,

when the circuit is closed, the terminal voltage of the battery,

V = E – Ir

i.e, V = 10 – (0.5) (3)

V = 10 – 1.5 = 8.5 V

or V = 1R; i.e., V = 0.5 × 17 = 8.5V

Question 21.

(a) Three resistors 2Ω, 4Ω and 5Ω are connected in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through, each resistor and the total current drawn from the battery.

Answer:

(a) Given R_{1} = 2Ω, R_{2} = 4Ω, R_{3} = 5Ω

(b) Main current

Question 22.

A negligbly small current is passed “through a wire of length 15m, uniform cross – section 6.0 × 10^{-7} m^{2} and its resistance is measured to be 5.0Ω. What is the resistivity of the material at the temperature of the experiment?

Answer:

Given, area of cross section A = 6 × 10^{-7} m^{2}

resistance R = 5.0Ω

resistivity p = ?

length of the wire 1 = 15 m,

We know that p = \(\frac{R A}{l}\)

i.e., p = \(\frac{5 \times 6 \times 10^{-7}}{15}\)

i.e., p = 2.0 × 10^{-7} Ωm.

Thus, the reistivity of the material is 2.0 × 10^{-7} Ωm.

Question 23.

A heating element usine nichrome connected to a 230V supply draws an initial current of 3.2A which settles after a few seconds to a steady value of 2.8A. What is the steady temperature of the heating element, if the room temperature is 27.0°C? Temperature coefficient of resistance of nichronie averaged over the teniperature range involved is 1.70 × 10^{-4}/°C.

Answer:

Given

Thus, temperature coefficient of resistance of nichrome = 871.1°C

Question 24.

(a) In a metre bridge, the balance point is found to at 37.5 cm from the end A, when the resistor y is of 12.5Ω Determine the resistance of x. Why are the connections between resistors in a Wheatstone or metre bridge made of thick copper strips?

(b) Determine the balance point of the bridge above, if x and y are interchanged.

(c) What happens, if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Answer:

(a) Given balance point from the end A, l = 39.5 cm

Resistance in the left gap x = ?

Resistance in the right gap y = 12.5Ω

Applying the balance condition of the metre bridge, we write \(\frac{x}{y}=\frac{l}{100-l}\)

i.e., x = 8.16Ω

Thick metal strips offer a low resistance. Hence, the power loss and the heat loss at the junctions are kept negligibly small or minimum.

(b) If x and y are interchanged, the ratio of balancing lengths also gets interchanged.

12.5 × 100 – 12.5l’ = 8.16l’

20.66 l’ = 1250.

or l’ = \(\frac{1250}{20.66}\) = 60.5 cm

Hence the balancing length changes to 60.5 cm.

(c) If the galvanometer and the cell were interchanged then the balancing length was not . going to be altered and the current in the galvanometer would be zero.

Question 25.

A storage battery of emf 8.0V and internal resistance 0.5Ω is being charged by a 120V DC supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer:

Given emf of a battery E’ = 8V

emf of a supply voltage V = 120 V

When the battery is being charged, its negative terminal is connected to the positive terminal of the DC source.

V_{AB} = V – IR

Where V_{AB} = E + r(1)

∴ E + IR = V – IR

i.e., 8 = 120 – (161)

where, R = 15.5 Ω, r = 0.5 Ω

∴ I = \(\frac{112}{16}\) = 7A

Terminal p.d. V_{AB} = E + (r)I = 8 + (0.5 × 7)

i.e., V_{AB} = 11.5V which is greater than the emf of a storage battery while charging.

Question 26.

In a potentiometer arrangement, a cell of emf 1.25V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by other cell the balance point shifts to 63.0 cm, what is the cmf of the second cell? Answer:

Givcn E = 1.25V, l = 35.Ocm

E’ = ? l = 163.0cm

We know that E α l in a potentiometer

∴ \(\frac{E^{\prime}}{E}=\frac{l^{\prime}}{l}\)

or E’ = \(\frac{E l^{\prime}}{l}=\frac{1.25 \times 63}{35}\)

i.e., E’ = 2.25 V

Thus, the emf of the other cell is 2.25 V

Question 27.

The number density of free electrons in a copper conductor is estimated as 8.5 × 10^{28} m^{-3}. How long does an electron take to drift from one end of a wire 3.0m long to its other end? The area of cross-section of the wire is 2.0 × 10^{-6}m^{2} and it is carrying a current of 3.0A.

Answer:

n = 8.5 × 10^{28} m^{-3}

I = 3.0 m

A = 2 × 10^{-6} m^{2}

I = 3.0 A

Current in a conductor I = n AeV_{d}

i.e., drift velocity V_{d} = 0.1103 × 10^{-3} ms^{-1}

The time taken for current to drift = 2.72 × 10^{4}s

Question 28.

The earth’s surface has a negative surface charge density of 10^{-9} Cm^{-2}. The potential difference of 400kV between the top of the atmosphere and the surface results in a current of only 1800A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time would be required to neutralize the earth’s surface charge? Take radius of earth = 6.37 × 10^{6} m.

Answer:

Given σ = 10^{-9} Cm^{-2}, V = 400 × 10^{3} V, I = 1800 A, t = ? , R = 6.37 × 106m

Surface area of the earth = 4πR^{2}

i.e., A = 4 × 3,142 × (6.37)^{2} × 10^{12} m^{2}

A = 5.0964 × 10^{14}m^{2}

Total charge on the earth Q = A

i.e., Q = 5.0964 × 10^{14} × 10^{-9}C

i.e., Q = 5.096 × 10^{5}C

Time required for total discharge

t = \(\frac{Q}{I}=\frac{5.096 \times 10^{5}}{1800}\)

i.e, , t = 2.83 × 10^{2}s

Thus, the time required to neutralize the earth’s surface charge is 2.83 × 10^{2}s.

Question 29.

(a) Six lead – acid type of secondary cells each of emf 2.0V and internal resistance 0.015Ω are joined in series to provide supply to a resistance of 8.5Ω. What is the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9V and large internal resistance of 380 Ω.

What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

Given emf of each cell = 2.0V

Total emf of six cells = 2 × 6 = 12V

Total internal resistance = 6 × 0.015Ω = 0.090 Ω

Load resistance = 8.5Ω

The terminal voltage of the battery = V = IR

i.e, V = 1.4 × 8.5 V

i.e., V = 11.9 V

(b) Given E = 1.9V, r = 380Ω

The current required to start the motor is about 2 × 10^{4} times the maximum current of 5 × 10^{-3}A. Hence the cell cannot be used to start the car.

Question 30.

The figure shows a, potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0Ω is found to be 58.3cm, while that with the unknown resistance ‘X’ is 68.5 cm. Determine the value of x. What might you do if you failed to find a balance point with the given cell of emf ε?

Answer:

Given : l_{1} = 53.3 cm, l_{2} = 68.5cm, R = 10Ω, X = ?

Let T be the current in the potentiometer wire.

Let V_{1} and V_{2} be potential drops across R and X respectively.

∴ x = 11.75 Ω

If we fail to find the balance point with the given cell of emf e then the p.d across R or X will be greater than the p.d. across the potentiometer wire. A suitable resistor of high resistance may be used in series with the cell of emf ‘s’.

Question 31.

The terminal potential difference across a load resistor of resistance 20Ω connected to an emf source of internal resistance 3Ω is 10V. Calculate the emf of the source. If an another resistor of resistance 30Ω is connected in series then calculate the new terminal potential difference across the cell. (March 2014)

Answer:

Case (i):

Terminal potential difference V = \(\frac{E R}{R+r}\)

i.e., 10 = \(\frac{E(20)}{20+3}\)

∴ E = 11.5V

Case (ii):

Question 32.

The figure shows a 2.0V potentiometer used for the determination of internal resistance of a 1.5V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer:

Given the balancing length when the cell is in the open circuit l_{1} = 76.3cm

The balancing length when the cell is in the closed circuit with a load resistor, l_{2} = 64.8 cm, and R = 9.5Ω

We know that the internal resistance

Hence internal resistance of the cell is 1.68Ω