Current Electricity Easy Questions and Answers | Page - 5

Current Electricity

In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is 0.367×10^–3 kg / C)

1. 1.76 × 10^–3 kg
1. 9.67 × 10^–3 kg
1. 17.61 × 10^–3 kg
1. 3.67 × 10^–3 kg

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I = P = 100 × 10³ A = 10⁵ A
V 125 60

E.C.E. = 0.367 × 10^-6 kg C ^-1
Charge per minute = (I × 60) C
= 10⁵ × 60 C = 6 × 10⁶ C
125 125

∴Mass liberated
= 6 × 10⁶ × 0.367 × 10^-6
125

= 6 × 1000 × 10^-3 = 17.616 × 10^-3kg
125

Correct Option: C

I = P = 100 × 10³ A = 10⁵ A
V 125 60

E.C.E. = 0.367 × 10^-6 kg C ^-1
Charge per minute = (I × 60) C
= 10⁵ × 60 C = 6 × 10⁶ C
125 125

∴Mass liberated
= 6 × 10⁶ × 0.367 × 10^-6
125

= 6 × 1000 × 10^-3 = 17.616 × 10^-3kg
125

A thermocouple of negligible resistance produces an e.m.f. of 40 µV/°C in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1µA/div, is employed with the termocouple. The smallest value of temperature difference that can be detected by the system will be

1. 0.5°C
1. 1°C
1. 0.1°C
1. 0.25°C

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1 division = 1µA
Current for 1°C = 40μV/10 = 4μA
1μA =1/4 °C = 0.25°C.

Correct Option: D

1 division = 1µA
Current for 1°C = 40μV/10 = 4μA
1μA =1/4 °C = 0.25°C.

In the circuit shown in the figure, if potential at point A is taken to be zero, the potential at point B is

1. –1V
1. + 2V
1. –2V
1. + 1V

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Current from D to C = 1A
∴ V_D – V_C = 2 × 1 = 2V
V_A = 0
∴ V_C = 1V,
∴ V_D – VC = 2
⇒ V_D – 1 = 2
∴ V_D = 3V
∴ V_D – V_B = 2
∴ 3 – V_B = 2 ∴ V_B = 1V

Correct Option: D

Current from D to C = 1A
∴ V_D – V_C = 2 × 1 = 2V
V_A = 0
∴ V_C = 1V,
∴ V_D – VC = 2
⇒ V_D – 1 = 2
∴ V_D = 3V
∴ V_D – V_B = 2
∴ 3 – V_B = 2 ∴ V_B = 1V

The current in the following circuit is

1. 1A
1. 2/3 A
1. 2/9 A
1. 1/8 A

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Resistance of ACB, R' = 3Ω + 3Ω = 6Ω.
For net resistance between A and B; R' = 6Ω and 3Ω are in parallel.
R_AB = 3 ×6 = 18 = 2 Ω
3 + 6 9

Current in circuit (I) = E = 2 = 1 A
R_AB 2

Correct Option: A

Resistance of ACB, R' = 3Ω + 3Ω = 6Ω.
For net resistance between A and B; R' = 6Ω and 3Ω are in parallel.
R_AB = 3 ×6 = 18 = 2 Ω
3 + 6 9

Current in circuit (I) = E = 2 = 1 A
R_AB 2

Consider the following two statements:
(A)Kirchhoff's junction law follows from the conservation of charge.
(B)Kirchhoff's loop law follows from the conservation of energy.
Which of the following is correct?

1. Both (A) and (B) are wrong
1. (a) is correct and (B) is wrong
1. (a) is wrong and (B) is correct
1. Both (A) and (B) are correct

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Junction law follows from conservation of charge and loop law is the conservation of energy

Correct Option: D

Junction law follows from conservation of charge and loop law is the conservation of energy