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Current Electricity

268493 The potential difference across the terminals of a battery is $10 V$ when there is a current of $3 A$ in the battery from the negative to the positive terminal. When the current is $2 A$ in the reverse direction, the potential difference becomes $15 V$ . T he internal resistance of the battery is

1

1 Ω

2

0.4 Ω

3

0.6 Ω

4

0.8 Ω

Explanation:

$E - 3 r = 10$
$E + 2 r = 15$
solving (i) \& (ii) $r = 1$

Current Electricity

268494 Two cells ofemf $3 V$ and $5 V$ and internal resistance $r_{1}$ and $r_{2}$ respectively are in series with an external resistance R. If the p.d. across 1st cell is zero, then $R$ is

1

\frac{5 r_{1} - 3 r_{2}}{3}

2

\frac{2 r_{1} - 3 r_{2}}{4}

3

\frac{3 r_{1} - 5 r_{2}}{3}

4

\frac{4 r_{1} - 5 r_{2}}{3}

Explanation:

$V = E$ - ir

Current Electricity

268495 A battery when connected by a resistance of $16 Ω$ gives a terminal voltage of $12 V$ . and when connected by a resistance of $10 Ω$ gives a terminal voltage of 11V. Then the emf of the battery and its internal resistance

1

12.8 V

2

13.7 V

3

10.7 V

4

9 V

Explanation:

$E = \frac{V}{R} (R + r) =$ constant

Current Electricity

268496 When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is $0.5 A$ . Instead, when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

6 Ω

Explanation:

$E = i (R + r)$

Current Electricity

268497 Two cells ofemf $4 V$ and $8 V$ are connected to two resister $4 Ω$ and $6 Ω$ as shown. If $8 V$ cell is short circuited. Then current through resistance $4 Ω$ and $6 Ω$

1 2 A

2 1 A

3 2.5 A

4 3 A

Explanation:

$V = i R$

Current Electricity

268493 The potential difference across the terminals of a battery is $10 V$ when there is a current of $3 A$ in the battery from the negative to the positive terminal. When the current is $2 A$ in the reverse direction, the potential difference becomes $15 V$ . T he internal resistance of the battery is

1

1 Ω

2

0.4 Ω

3

0.6 Ω

4

0.8 Ω

Explanation:

$E - 3 r = 10$
$E + 2 r = 15$
solving (i) \& (ii) $r = 1$

Current Electricity

268494 Two cells ofemf $3 V$ and $5 V$ and internal resistance $r_{1}$ and $r_{2}$ respectively are in series with an external resistance R. If the p.d. across 1st cell is zero, then $R$ is

1

\frac{5 r_{1} - 3 r_{2}}{3}

2

\frac{2 r_{1} - 3 r_{2}}{4}

3

\frac{3 r_{1} - 5 r_{2}}{3}

4

\frac{4 r_{1} - 5 r_{2}}{3}

Explanation:

$V = E$ - ir

Current Electricity

268495 A battery when connected by a resistance of $16 Ω$ gives a terminal voltage of $12 V$ . and when connected by a resistance of $10 Ω$ gives a terminal voltage of 11V. Then the emf of the battery and its internal resistance

1

12.8 V

2

13.7 V

3

10.7 V

4

9 V

Explanation:

$E = \frac{V}{R} (R + r) =$ constant

Current Electricity

268496 When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is $0.5 A$ . Instead, when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

6 Ω

Explanation:

$E = i (R + r)$

Current Electricity

268497 Two cells ofemf $4 V$ and $8 V$ are connected to two resister $4 Ω$ and $6 Ω$ as shown. If $8 V$ cell is short circuited. Then current through resistance $4 Ω$ and $6 Ω$

1 2 A

2 1 A

3 2.5 A

4 3 A

Explanation:

$V = i R$

Current Electricity

268493 The potential difference across the terminals of a battery is $10 V$ when there is a current of $3 A$ in the battery from the negative to the positive terminal. When the current is $2 A$ in the reverse direction, the potential difference becomes $15 V$ . T he internal resistance of the battery is

1

1 Ω

2

0.4 Ω

3

0.6 Ω

4

0.8 Ω

Explanation:

$E - 3 r = 10$
$E + 2 r = 15$
solving (i) \& (ii) $r = 1$

Current Electricity

268494 Two cells ofemf $3 V$ and $5 V$ and internal resistance $r_{1}$ and $r_{2}$ respectively are in series with an external resistance R. If the p.d. across 1st cell is zero, then $R$ is

1

\frac{5 r_{1} - 3 r_{2}}{3}

2

\frac{2 r_{1} - 3 r_{2}}{4}

3

\frac{3 r_{1} - 5 r_{2}}{3}

4

\frac{4 r_{1} - 5 r_{2}}{3}

Explanation:

$V = E$ - ir

Current Electricity

268495 A battery when connected by a resistance of $16 Ω$ gives a terminal voltage of $12 V$ . and when connected by a resistance of $10 Ω$ gives a terminal voltage of 11V. Then the emf of the battery and its internal resistance

1

12.8 V

2

13.7 V

3

10.7 V

4

9 V

Explanation:

$E = \frac{V}{R} (R + r) =$ constant

Current Electricity

268496 When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is $0.5 A$ . Instead, when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

6 Ω

Explanation:

$E = i (R + r)$

Current Electricity

268497 Two cells ofemf $4 V$ and $8 V$ are connected to two resister $4 Ω$ and $6 Ω$ as shown. If $8 V$ cell is short circuited. Then current through resistance $4 Ω$ and $6 Ω$

1 2 A

2 1 A

3 2.5 A

4 3 A

Explanation:

$V = i R$

Current Electricity

268493 The potential difference across the terminals of a battery is $10 V$ when there is a current of $3 A$ in the battery from the negative to the positive terminal. When the current is $2 A$ in the reverse direction, the potential difference becomes $15 V$ . T he internal resistance of the battery is

1

1 Ω

2

0.4 Ω

3

0.6 Ω

4

0.8 Ω

Explanation:

$E - 3 r = 10$
$E + 2 r = 15$
solving (i) \& (ii) $r = 1$

Current Electricity

268494 Two cells ofemf $3 V$ and $5 V$ and internal resistance $r_{1}$ and $r_{2}$ respectively are in series with an external resistance R. If the p.d. across 1st cell is zero, then $R$ is

1

\frac{5 r_{1} - 3 r_{2}}{3}

2

\frac{2 r_{1} - 3 r_{2}}{4}

3

\frac{3 r_{1} - 5 r_{2}}{3}

4

\frac{4 r_{1} - 5 r_{2}}{3}

Explanation:

$V = E$ - ir

Current Electricity

268495 A battery when connected by a resistance of $16 Ω$ gives a terminal voltage of $12 V$ . and when connected by a resistance of $10 Ω$ gives a terminal voltage of 11V. Then the emf of the battery and its internal resistance

1

12.8 V

2

13.7 V

3

10.7 V

4

9 V

Explanation:

$E = \frac{V}{R} (R + r) =$ constant

Current Electricity

268496 When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is $0.5 A$ . Instead, when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

6 Ω

Explanation:

$E = i (R + r)$

Current Electricity

268497 Two cells ofemf $4 V$ and $8 V$ are connected to two resister $4 Ω$ and $6 Ω$ as shown. If $8 V$ cell is short circuited. Then current through resistance $4 Ω$ and $6 Ω$

1 2 A

2 1 A

3 2.5 A

4 3 A

Explanation:

$V = i R$

Current Electricity

268493 The potential difference across the terminals of a battery is $10 V$ when there is a current of $3 A$ in the battery from the negative to the positive terminal. When the current is $2 A$ in the reverse direction, the potential difference becomes $15 V$ . T he internal resistance of the battery is

1

1 Ω

2

0.4 Ω

3

0.6 Ω

4

0.8 Ω

Explanation:

$E - 3 r = 10$
$E + 2 r = 15$
solving (i) \& (ii) $r = 1$

Current Electricity

268494 Two cells ofemf $3 V$ and $5 V$ and internal resistance $r_{1}$ and $r_{2}$ respectively are in series with an external resistance R. If the p.d. across 1st cell is zero, then $R$ is

1

\frac{5 r_{1} - 3 r_{2}}{3}

2

\frac{2 r_{1} - 3 r_{2}}{4}

3

\frac{3 r_{1} - 5 r_{2}}{3}

4

\frac{4 r_{1} - 5 r_{2}}{3}

Explanation:

$V = E$ - ir

Current Electricity

268495 A battery when connected by a resistance of $16 Ω$ gives a terminal voltage of $12 V$ . and when connected by a resistance of $10 Ω$ gives a terminal voltage of 11V. Then the emf of the battery and its internal resistance

1

12.8 V

2

13.7 V

3

10.7 V

4

9 V

Explanation:

$E = \frac{V}{R} (R + r) =$ constant

Current Electricity

268496 When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is $0.5 A$ . Instead, when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

6 Ω

Explanation:

$E = i (R + r)$

Current Electricity

268497 Two cells ofemf $4 V$ and $8 V$ are connected to two resister $4 Ω$ and $6 Ω$ as shown. If $8 V$ cell is short circuited. Then current through resistance $4 Ω$ and $6 Ω$

1 2 A

2 1 A

3 2.5 A

4 3 A

Explanation:

$V = i R$