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Current Electricity

268453 A $5 V$ battery with internal resistance $2 Ω$ and a $2 V$ battery with internal resistance $1 Ω$ are connected in parallel with unlike polarities connected together .This combination is connected to $10 Ω$ resistor The current in the $10 Ω$ resistor is

1

0.27 A

2

0.05 A

3

0.25 A

4

0.3 A

Explanation:

$I_{s} = \frac{2 E}{2 r + R}$ , and $I_{P} = \frac{2 E}{r + 2 R}$

Current Electricity

268454 A voltmeter resistance $500 Ω$ is used to measure the emf of a cell of internal resistance $4 Ω$ . The percentage error in the reading of the voltmeter will be

1

0.4 %

2

0.6 %

3

0.8 %

4

1.2 %

Explanation:

$%$ error $= \frac{Δ E}{E} \times 100 = \frac{i r}{E} \times 100$

Current Electricity

268455 When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is

1

4 Ω

2

2 Ω

3

1 Ω

4

7 Ω

Explanation:

when $i_{s} = i_{p}; r = R$

Current Electricity

268456 Two cells with same e.m.f. ' $E$ ' and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance ' $R$ '. The value of $R$ so that the p.d. across the first cell be zero is

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1} + r_{2}

4

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

Explanation:

$i = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}; V_{1} = E_{1} - i r_{1}$

Current Electricity

268453 A $5 V$ battery with internal resistance $2 Ω$ and a $2 V$ battery with internal resistance $1 Ω$ are connected in parallel with unlike polarities connected together .This combination is connected to $10 Ω$ resistor The current in the $10 Ω$ resistor is

1

0.27 A

2

0.05 A

3

0.25 A

4

0.3 A

Explanation:

$I_{s} = \frac{2 E}{2 r + R}$ , and $I_{P} = \frac{2 E}{r + 2 R}$

Current Electricity

268454 A voltmeter resistance $500 Ω$ is used to measure the emf of a cell of internal resistance $4 Ω$ . The percentage error in the reading of the voltmeter will be

1

0.4 %

2

0.6 %

3

0.8 %

4

1.2 %

Explanation:

$%$ error $= \frac{Δ E}{E} \times 100 = \frac{i r}{E} \times 100$

Current Electricity

268455 When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is

1

4 Ω

2

2 Ω

3

1 Ω

4

7 Ω

Explanation:

when $i_{s} = i_{p}; r = R$

Current Electricity

268456 Two cells with same e.m.f. ' $E$ ' and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance ' $R$ '. The value of $R$ so that the p.d. across the first cell be zero is

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1} + r_{2}

4

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

Explanation:

$i = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}; V_{1} = E_{1} - i r_{1}$

Current Electricity

268453 A $5 V$ battery with internal resistance $2 Ω$ and a $2 V$ battery with internal resistance $1 Ω$ are connected in parallel with unlike polarities connected together .This combination is connected to $10 Ω$ resistor The current in the $10 Ω$ resistor is

1

0.27 A

2

0.05 A

3

0.25 A

4

0.3 A

Explanation:

$I_{s} = \frac{2 E}{2 r + R}$ , and $I_{P} = \frac{2 E}{r + 2 R}$

Current Electricity

268454 A voltmeter resistance $500 Ω$ is used to measure the emf of a cell of internal resistance $4 Ω$ . The percentage error in the reading of the voltmeter will be

1

0.4 %

2

0.6 %

3

0.8 %

4

1.2 %

Explanation:

$%$ error $= \frac{Δ E}{E} \times 100 = \frac{i r}{E} \times 100$

Current Electricity

268455 When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is

1

4 Ω

2

2 Ω

3

1 Ω

4

7 Ω

Explanation:

when $i_{s} = i_{p}; r = R$

Current Electricity

268456 Two cells with same e.m.f. ' $E$ ' and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance ' $R$ '. The value of $R$ so that the p.d. across the first cell be zero is

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1} + r_{2}

4

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

Explanation:

$i = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}; V_{1} = E_{1} - i r_{1}$

Current Electricity

268453 A $5 V$ battery with internal resistance $2 Ω$ and a $2 V$ battery with internal resistance $1 Ω$ are connected in parallel with unlike polarities connected together .This combination is connected to $10 Ω$ resistor The current in the $10 Ω$ resistor is

1

0.27 A

2

0.05 A

3

0.25 A

4

0.3 A

Explanation:

$I_{s} = \frac{2 E}{2 r + R}$ , and $I_{P} = \frac{2 E}{r + 2 R}$

Current Electricity

268454 A voltmeter resistance $500 Ω$ is used to measure the emf of a cell of internal resistance $4 Ω$ . The percentage error in the reading of the voltmeter will be

1

0.4 %

2

0.6 %

3

0.8 %

4

1.2 %

Explanation:

$%$ error $= \frac{Δ E}{E} \times 100 = \frac{i r}{E} \times 100$

Current Electricity

268455 When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is

1

4 Ω

2

2 Ω

3

1 Ω

4

7 Ω

Explanation:

when $i_{s} = i_{p}; r = R$

Current Electricity

268456 Two cells with same e.m.f. ' $E$ ' and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance ' $R$ '. The value of $R$ so that the p.d. across the first cell be zero is

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1} + r_{2}

4

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

Explanation:

$i = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}; V_{1} = E_{1} - i r_{1}$