Current Electricity Important Questions

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Important Questions on Current Electricity

Ques 1. Why is a potentiometer preferred over a voltmeter for determining the emf of a cell? (2 Marks)

Ans. Potentiometer is preferred over a voltmeter for measuring emf because a potentiometer only measures the emf of the cell and does not get influenced by the current. Whereas, a voltmeter draws some current and measures the terminal voltage of a cell. 

Ques 2. Two electric bulbs P and Q have their resistances in the ratio 1:2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. (2 Marks)

Ans. Given, R_P / R_Q = 1/2

Power Dissipated = P_P / P_Q = I2R_P / I2R_Q = 1/2

Ques 3. The plot of the variation of potential difference A across a combination of three identical cells in series, versus current is shown along the question. What is the emf and internal resistance of each cell? (3 Marks)

Ans. In this question, 3 identical cells are connected in series, so,

E’ = E₁ + E₂ + E₃ = 3E = 6V (when I = 0)

Therefore, E = 2V 

Now, E = IR + V

IR = E - V

R = E - V / I

R = 2 - 0 / 1

R = 2Ω

Ques 4. Calculate the current drawn from the battery in the given network. (3 Marks)

Ans. Since R₅ / R₁ = R₃ / R₄ 

So, the current through R₂ = 5Ω is zero

Now, (R₁ , R₅) and (R₃ , R₄) are in series to give 3Ω and 6Ω respectively. They are together in parallel to give, 

R_eq = 3 × 6 / 3 + 6 = 18/9 = 2Ω

∴ Current drawn from the battery,

I = V / R_eq = 4 / 2 = 2 A

Ques 5. A wire of 15 Ω resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery. (3 Marks)

Ans. R = 15Ω

On stretching to double its original length, the resistance becomes R₁ = 60Ω, as on stretching volume is constant and R ∝ l2.

The two cut parts are connected in parallel, so their resistance is 30Ω each. 

R_eq = 30 / 2 = 15Ω

Current drawn = I = V / R_eq

I = 3 / 15 = 1 / 5 = 0.2 A

Ques 6. In the meter bridge experiment, balance point was observed at J with AJ = l. (3 Marks)

1. The values of R and X were doubled and then interchanged. What would be the new position of balance point?
2. If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected?

Ans. i. Balance point will change from I to (100 - I),

R / X = l / 100 - l 

2X / 2R = X / R = 100 - l / l

2. From the principle of Wheatstone Bridge,

R / X = l / 100 - l 

X = R (100 - l / l)

Ques 7. A battery of emf 10 V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, find

(i) the resistance of the resistor;

(ii) the terminal voltage of the battery. (3 Marks)

Ans. (i) I = V / r + R

So, 10 / r + R = 0.5

or 10 / 3 + R = 0.5

= 100 / 5 = 3 + R

∴ R = 20 - 3 = 17Ω 

(ii) V = IR

∴ V = 5 / 10 × 17 = 85 / 10 = 8.5 V

Ques 8. Define the terms:

(i) drift velocity,

(ii) relaxation time.

A conductor of length L is connected to a dc source of emf e. If this conductor is replaced by another conductor of the same material and same area of cross-section but of length 3L, how will the drift velocity change? (3 Marks)

Ans. (i) Drift Velocity: It may be defined as the average velocity gained by the free electrons of a conductor in the opposite direction of the externally applied field.

(ii) Relaxation Time: The average time that elapses between two successive collisions of an electron is called relaxation time.

V’_d = eV / m3L = ⅓ Vd 

so, V_d = eV / mL

When length is tripled (3L), drift velocity becomes one-third of the original.

Ques 9. The network PQRS, shown in the circuit diagram, has the batteries of 4 V and 5 V and negligible internal resistance. A milliammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter. (5 Marks)

Ans. Applying loop rule to loop PQRP

-4 = 60 (I – I₁) – 20 I₁ = 0

or – 4 = 60I – 60I₁ – 20I₁

or 20I₁ – 15 I = 1 …[+ by 4 …(i)]

Applying loop Yule to loop PRSP, we get

-5 + 200 I + 20 I₁ = 0

4I₁ + 40 I = 1 …[+ by 5 …(ii)]

Solving (i) and (ii), we get,

I₁ = 1₁ / 172 = 0.064 A

∴ Reading of Ammeter = 0.064A

Ques 10. Calculate the current drawn from the battery by the network of resistors shown in the figure. (5 Marks)

Ans. Rearrange the circuit diagram to form a Wheatstone Bridge, 

P/Q = R/S

1/2 = 2/4 = 1/2

It is the condition of null point when no current flows through BD (5Ω).

Resistance P (1Ω) and Q (2Ω) are in series;

Similarly, resistance Q = 2Ω and S are in series

R₂ = 2 + 4 = 6Ω

Now, R₁ and R₂ are in parallel,

1/R = 1/R₁ + 1/R₂ = 1/3 + 1/6 = ½

R = 2Ω

I = V/R = 4/2 = 2A 

∴ Current in the circuit is 2A.

Ques 11. State Kirchhoff’s rules. Use these rules to write the expressions for the current I1 , I2 and I3 in the circuit diagram shown. (5 Marks)

Ans. (i). Kirchhoff’s Junction Rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.

(ii) Kirchhoff’s Loop Rule: The algebraic sum of changes in potential-in any closed loop involving resistors and cells is zero.

Now, according to Kirchhoff’s Junction Rule, 

I₃ = I₁ + I₂ ……(i)

Considering loop FCDEF, 

3I₂ - 4I₂ = 1 ……(ii)

Considering loop FCBAF, 

3I₂ + 2I₃ = 3 ……(iii)

3I₂ + 2(I₁ + I₂) = 3

5I₂ + 2I₁ = 3 …….(iv)

Solving equations (i), (ii) and (iv), we get

I₁ = 2/13 A

I₂ = 7/13 A

I₃ = 9/13 A

Ques 12. In the circuit shown, R₁ = 4Ω, R₂ = R₃ = 15 Ω, R₄ = 30Ω and E = 10V. Calculate the equivalent resistance of the circuit and the current in each resistor. (5 Marks)

Ans. In the above figure, 

R₃ and R₄ are in parallel combination, so their resultant will be,

1/R₃₄ = 1/R₃ + 1/R₄ = 1/15 + 1/30 = 2 + 1 / 30 = 3/30 = 1/10

1/R₃₄ = 10Ω

Now, R₃₄ and R₂ are again in parallel combination;

1/R₂₃₄ = 1/R₃₄ + 1/R₂ = 1/10 + 1/15

= 15 + 10 / 150 = 25 / 150 = 1/6

R₂₃₄ = 6Ω

Now, R₂₃₄ and R₁ are in series combination,

R₁₂₃₄ = R₂₃₄ + R₁ = 6 + 4 = 10Ω

I = V/R = 10/10 = 1 A

∴ I₁ = 1 A (I₁ = I₂ + I₃ + I4)

Potential at R1, 

V₁ = I₁R₁ = 1 × 4 = 4V

Potential at R₂, R₃ and R₄, V₂₃₄ = 1 × 6 = 6 V

∴ I₂ = 6/15 = ⅖ A 

I₃ = 6/15 = ⅖ A

I₄ = 6/30 = ⅕ A 

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