Current Electricity Easy Questions and Answers | Page - 3

Current Electricity

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is

1. 0.5 Ω
1. 0.8 Ω
1. 1.0 Ω
1. 0.2 Ω

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Given : emf ε = 2.1 V
I = 0.2 A, R = 10Ω
Internal resistance r = ?
From formula.
ε – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2 r = 2 or 0.2 r = 0.1
⇒ r = 0.1 = 0.5 Ω
0.2

Correct Option: A

Given : emf ε = 2.1 V
I = 0.2 A, R = 10Ω
Internal resistance r = ?
From formula.
ε – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2 r = 2 or 0.2 r = 0.1
⇒ r = 0.1 = 0.5 Ω
0.2

Cell having an emf ε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by :

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The current through the resistance R
I = R
ε
R + r

The potential difference across R
V = IR = R
ε
R + r

when R = 0, V = 0,
R = ∞, V = ε Thus V increases as R increases upto certain limit, but it does not increase further.

Correct Option: C

The current through the resistance R
I = R
ε
R + r

The potential difference across R
V = IR = R
ε
R + r

when R = 0, V = 0,
R = ∞, V = ε Thus V increases as R increases upto certain limit, but it does not increase further.

A current of 2A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9Ω resistor. The internal resistance of the battery is

1. 0.5 Ω
1. 1/3 Ω
1. 1/4 Ω
1. 1 Ω

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Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by
i = E
R + r

In first case,
2 = E ...(i)
2 + r

In second case,
0.5 = E ...(i)
9 + r

From (1) & (2), 4 + 2r = 4.5 + 0.5 r
⇒1.5 r = 0.5 ⇒ r = 1/3 Ω.

Correct Option: B

Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by
i = E
R + r

In first case,
2 = E ...(i)
2 + r

In second case,
0.5 = E ...(i)
9 + r

From (1) & (2), 4 + 2r = 4.5 + 0.5 r
⇒1.5 r = 0.5 ⇒ r = 1/3 Ω.

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is

1. 1.0 ohm
1. 0.5 ohm
1. 2.0 ohm
1. zero

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Here
E > ER
R + r

hence the lengths 110 cm and 100 cm are interchanged.
Without being short-circuited through R, only the battery E is balanced.
E = V × l₁ = V × 110 ................(i)
L L

When R is connected across E, Ri = V × l₂
L

Or, R E = V × 100..............(ii)
R + r L

Dividing (i) by (ii), we get
R + r = 110
R 100

or, 100 R + 100 r = 110 R

Correct Option: A

Here
E > ER
R + r

hence the lengths 110 cm and 100 cm are interchanged.
Without being short-circuited through R, only the battery E is balanced.
E = V × l₁ = V × 110 ................(i)
L L

When R is connected across E, Ri = V × l₂
L

Or, R E = V × 100..............(ii)
R + r L

Dividing (i) by (ii), we get
R + r = 110
R 100

or, 100 R + 100 r = 110 R

A steady current of 1.5 amp flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of copper is 30 × 10^–5 g coulomb^–1, the mass of copper deposited on the electrode will be

1. 0.50 g
1. 0.67 g
1. 0.27 g
1. 0.40 g.

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We have, m = ZIt
where, Z is the electrochemical equivalent of copper.
⇒ m = 30 x 10^-5 x 1.5 x 10 x 60
= 0.27 gm.

Correct Option: C

We have, m = ZIt
where, Z is the electrochemical equivalent of copper.
⇒ m = 30 x 10^-5 x 1.5 x 10 x 60
= 0.27 gm.