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Current Electricity

152500 A fuse wire of radius $0.2 mm$ blows off with a current of $5 A$ . The fuse wire of same material but of radius $0.3 mm$ will blow off with a current of

1

\frac{15}{2} A

2

\frac{5 \sqrt{3}}{2} A

3

5 \sqrt{\frac{27}{8}} A

4

5 A

Explanation:

C Given, $r_{1} = 0.2 mm$
$r_{2} = 0.3 mm$
Current $(i_{1}) = 5 A$
We know
$i^{2} \propto r^{3}$
$\frac{i_{1}}{i_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2} = {(\frac{0.2}{0.3})}^{3 / 2}$
$i_{2} = 5 \sqrt{\frac{27}{8}} A$

AP EAMCET (20.04.2019) Shift-1

Current Electricity

152502 Six similar bulbs are connected as shown in the figure with a DC source of emf $E$ and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section $A$ and one from section $B$ are glowing, will be

1

9 : 4

2

1 : 2

3

2 : 1

4

4 : 9

Explanation:

A Case- I : When all bulb are glowing then

$R_{eq} = \frac{R}{3} + \frac{R}{3} = \frac{2 R}{3}$
Then power $(P_{1}) = \frac{E^{2}}{R_{eq}} = \frac{3 E^{2}}{2 R}$
Case -II : When two from section $A$ and one from section $B$ are glowing

$R_{eq} = \frac{R}{2} + R = \frac{3 R}{2}$
Then, power $(P_{2}) = \frac{2 E^{2}}{3 R}$
Then $\frac{P_{1}}{P_{2}} = \frac{3 E^{2} \times 3 R}{2 R \times 2 E^{2}} = 9 : 4$

NEET National-2019

Current Electricity

152503 An inverter battery operates on $24 V$ and has negligible internal resistance. It is rated at 140 ampere-hour. What external resistance would have to be connected to the battery if it were to be discharged in 14 hours

1

1.6 Ω

2

2.4 Ω

3

5.9 Ω

4

10.0 Ω

Explanation:

B Given that,
Potential of battery $(V) = 24 V$
Internal Resistance $(r) = 0$
Rate of battery $(Q) = 140$ ampere - hour
$= 140 \times 3600 A - s$
$= 5.04 \times 10^{5} C$
Time taken to discharge battery $(t) = 14 hr$ If the external resistance is $R$ -
Then,
$V = IR$
$R = \frac{V}{I}$
$R = \frac{V . t}{Q}$
$= \frac{24 \times t}{5.04 \times 10^{5}}$
$[∵ I = Q / t]$
$∵ t = 14 \times 3600 \sec$
$∵ R = \frac{24 \times 14 \times 3600}{5.04 \times 10^{5}}$
$R = 2.4 Ω$

AMU-2018

Current Electricity

152504 Two identical cells each of emf $E$ and internal resistance $r$ are connected in parallel with an external resistance $R$ . To get maximum power developed across $R$ , the value of $R$ is

1

R = r / 2

2

R = r

3

R = r / 3

4

R = 2 r

Explanation:

A
For maximum power
$R_{internal} = R_{load}$
$R_{int} = \frac{r \times r}{r + r} = \frac{r}{2}$
$R = \frac{r}{2}$
Hence the maximum power developed across, $R =$ $(\frac{r}{2})$

Current Electricity

152500 A fuse wire of radius $0.2 mm$ blows off with a current of $5 A$ . The fuse wire of same material but of radius $0.3 mm$ will blow off with a current of

1

\frac{15}{2} A

2

\frac{5 \sqrt{3}}{2} A

3

5 \sqrt{\frac{27}{8}} A

4

5 A

Explanation:

C Given, $r_{1} = 0.2 mm$
$r_{2} = 0.3 mm$
Current $(i_{1}) = 5 A$
We know
$i^{2} \propto r^{3}$
$\frac{i_{1}}{i_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2} = {(\frac{0.2}{0.3})}^{3 / 2}$
$i_{2} = 5 \sqrt{\frac{27}{8}} A$

AP EAMCET (20.04.2019) Shift-1

Current Electricity

152502 Six similar bulbs are connected as shown in the figure with a DC source of emf $E$ and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section $A$ and one from section $B$ are glowing, will be

1

9 : 4

2

1 : 2

3

2 : 1

4

4 : 9

Explanation:

A Case- I : When all bulb are glowing then

$R_{eq} = \frac{R}{3} + \frac{R}{3} = \frac{2 R}{3}$
Then power $(P_{1}) = \frac{E^{2}}{R_{eq}} = \frac{3 E^{2}}{2 R}$
Case -II : When two from section $A$ and one from section $B$ are glowing

$R_{eq} = \frac{R}{2} + R = \frac{3 R}{2}$
Then, power $(P_{2}) = \frac{2 E^{2}}{3 R}$
Then $\frac{P_{1}}{P_{2}} = \frac{3 E^{2} \times 3 R}{2 R \times 2 E^{2}} = 9 : 4$

NEET National-2019

Current Electricity

152503 An inverter battery operates on $24 V$ and has negligible internal resistance. It is rated at 140 ampere-hour. What external resistance would have to be connected to the battery if it were to be discharged in 14 hours

1

1.6 Ω

2

2.4 Ω

3

5.9 Ω

4

10.0 Ω

Explanation:

B Given that,
Potential of battery $(V) = 24 V$
Internal Resistance $(r) = 0$
Rate of battery $(Q) = 140$ ampere - hour
$= 140 \times 3600 A - s$
$= 5.04 \times 10^{5} C$
Time taken to discharge battery $(t) = 14 hr$ If the external resistance is $R$ -
Then,
$V = IR$
$R = \frac{V}{I}$
$R = \frac{V . t}{Q}$
$= \frac{24 \times t}{5.04 \times 10^{5}}$
$[∵ I = Q / t]$
$∵ t = 14 \times 3600 \sec$
$∵ R = \frac{24 \times 14 \times 3600}{5.04 \times 10^{5}}$
$R = 2.4 Ω$

AMU-2018

Current Electricity

152504 Two identical cells each of emf $E$ and internal resistance $r$ are connected in parallel with an external resistance $R$ . To get maximum power developed across $R$ , the value of $R$ is

1

R = r / 2

2

R = r

3

R = r / 3

4

R = 2 r

Explanation:

A
For maximum power
$R_{internal} = R_{load}$
$R_{int} = \frac{r \times r}{r + r} = \frac{r}{2}$
$R = \frac{r}{2}$
Hence the maximum power developed across, $R =$ $(\frac{r}{2})$

Current Electricity

152500 A fuse wire of radius $0.2 mm$ blows off with a current of $5 A$ . The fuse wire of same material but of radius $0.3 mm$ will blow off with a current of

1

\frac{15}{2} A

2

\frac{5 \sqrt{3}}{2} A

3

5 \sqrt{\frac{27}{8}} A

4

5 A

Explanation:

C Given, $r_{1} = 0.2 mm$
$r_{2} = 0.3 mm$
Current $(i_{1}) = 5 A$
We know
$i^{2} \propto r^{3}$
$\frac{i_{1}}{i_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2} = {(\frac{0.2}{0.3})}^{3 / 2}$
$i_{2} = 5 \sqrt{\frac{27}{8}} A$

AP EAMCET (20.04.2019) Shift-1

Current Electricity

152502 Six similar bulbs are connected as shown in the figure with a DC source of emf $E$ and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section $A$ and one from section $B$ are glowing, will be

1

9 : 4

2

1 : 2

3

2 : 1

4

4 : 9

Explanation:

A Case- I : When all bulb are glowing then

$R_{eq} = \frac{R}{3} + \frac{R}{3} = \frac{2 R}{3}$
Then power $(P_{1}) = \frac{E^{2}}{R_{eq}} = \frac{3 E^{2}}{2 R}$
Case -II : When two from section $A$ and one from section $B$ are glowing

$R_{eq} = \frac{R}{2} + R = \frac{3 R}{2}$
Then, power $(P_{2}) = \frac{2 E^{2}}{3 R}$
Then $\frac{P_{1}}{P_{2}} = \frac{3 E^{2} \times 3 R}{2 R \times 2 E^{2}} = 9 : 4$

NEET National-2019

Current Electricity

152503 An inverter battery operates on $24 V$ and has negligible internal resistance. It is rated at 140 ampere-hour. What external resistance would have to be connected to the battery if it were to be discharged in 14 hours

1

1.6 Ω

2

2.4 Ω

3

5.9 Ω

4

10.0 Ω

Explanation:

B Given that,
Potential of battery $(V) = 24 V$
Internal Resistance $(r) = 0$
Rate of battery $(Q) = 140$ ampere - hour
$= 140 \times 3600 A - s$
$= 5.04 \times 10^{5} C$
Time taken to discharge battery $(t) = 14 hr$ If the external resistance is $R$ -
Then,
$V = IR$
$R = \frac{V}{I}$
$R = \frac{V . t}{Q}$
$= \frac{24 \times t}{5.04 \times 10^{5}}$
$[∵ I = Q / t]$
$∵ t = 14 \times 3600 \sec$
$∵ R = \frac{24 \times 14 \times 3600}{5.04 \times 10^{5}}$
$R = 2.4 Ω$

AMU-2018

Current Electricity

152504 Two identical cells each of emf $E$ and internal resistance $r$ are connected in parallel with an external resistance $R$ . To get maximum power developed across $R$ , the value of $R$ is

1

R = r / 2

2

R = r

3

R = r / 3

4

R = 2 r

Explanation:

A
For maximum power
$R_{internal} = R_{load}$
$R_{int} = \frac{r \times r}{r + r} = \frac{r}{2}$
$R = \frac{r}{2}$
Hence the maximum power developed across, $R =$ $(\frac{r}{2})$

Current Electricity

152500 A fuse wire of radius $0.2 mm$ blows off with a current of $5 A$ . The fuse wire of same material but of radius $0.3 mm$ will blow off with a current of

1

\frac{15}{2} A

2

\frac{5 \sqrt{3}}{2} A

3

5 \sqrt{\frac{27}{8}} A

4

5 A

Explanation:

C Given, $r_{1} = 0.2 mm$
$r_{2} = 0.3 mm$
Current $(i_{1}) = 5 A$
We know
$i^{2} \propto r^{3}$
$\frac{i_{1}}{i_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2} = {(\frac{0.2}{0.3})}^{3 / 2}$
$i_{2} = 5 \sqrt{\frac{27}{8}} A$

AP EAMCET (20.04.2019) Shift-1

Current Electricity

152502 Six similar bulbs are connected as shown in the figure with a DC source of emf $E$ and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section $A$ and one from section $B$ are glowing, will be

1

9 : 4

2

1 : 2

3

2 : 1

4

4 : 9

Explanation:

A Case- I : When all bulb are glowing then

$R_{eq} = \frac{R}{3} + \frac{R}{3} = \frac{2 R}{3}$
Then power $(P_{1}) = \frac{E^{2}}{R_{eq}} = \frac{3 E^{2}}{2 R}$
Case -II : When two from section $A$ and one from section $B$ are glowing

$R_{eq} = \frac{R}{2} + R = \frac{3 R}{2}$
Then, power $(P_{2}) = \frac{2 E^{2}}{3 R}$
Then $\frac{P_{1}}{P_{2}} = \frac{3 E^{2} \times 3 R}{2 R \times 2 E^{2}} = 9 : 4$

NEET National-2019

Current Electricity

152503 An inverter battery operates on $24 V$ and has negligible internal resistance. It is rated at 140 ampere-hour. What external resistance would have to be connected to the battery if it were to be discharged in 14 hours

1

1.6 Ω

2

2.4 Ω

3

5.9 Ω

4

10.0 Ω

Explanation:

B Given that,
Potential of battery $(V) = 24 V$
Internal Resistance $(r) = 0$
Rate of battery $(Q) = 140$ ampere - hour
$= 140 \times 3600 A - s$
$= 5.04 \times 10^{5} C$
Time taken to discharge battery $(t) = 14 hr$ If the external resistance is $R$ -
Then,
$V = IR$
$R = \frac{V}{I}$
$R = \frac{V . t}{Q}$
$= \frac{24 \times t}{5.04 \times 10^{5}}$
$[∵ I = Q / t]$
$∵ t = 14 \times 3600 \sec$
$∵ R = \frac{24 \times 14 \times 3600}{5.04 \times 10^{5}}$
$R = 2.4 Ω$

AMU-2018

Current Electricity

152504 Two identical cells each of emf $E$ and internal resistance $r$ are connected in parallel with an external resistance $R$ . To get maximum power developed across $R$ , the value of $R$ is

1

R = r / 2

2

R = r

3

R = r / 3

4

R = 2 r

Explanation:

A
For maximum power
$R_{internal} = R_{load}$
$R_{int} = \frac{r \times r}{r + r} = \frac{r}{2}$
$R = \frac{r}{2}$
Hence the maximum power developed across, $R =$ $(\frac{r}{2})$

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