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PHXII03:CURRENT ELECTRICITY

357146 In the circuit shown in figure, the heat produced in 5 $o h m$ resistance is 10 calories per second. The heat produced in $4 Ω$ resistance is
supporting img

1

3 c a l / \sec

2

4 c a l / \sec

3

1 c a l / \sec

4

2 c a l / \sec

Explanation:

Let $i_{1}$ and $i_{2}$ be the currents through the resistors.
$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{10}{5} = \frac{2}{1}$
Also heat produced per sec is, $P = i^{2} R$
$\Rightarrow \frac{P_{5}}{P_{4}} = {(\frac{i_{1}}{i_{2}})}^{2} \times \frac{5}{4} = {(\frac{2}{1})}^{2} \times \frac{5}{4} = \frac{5}{1}$
$\Rightarrow P_{4} = \frac{10}{5} = 2 c a l / \sec$
supporting img

PHXII03:CURRENT ELECTRICITY

357147 Two heater wires of equal lengths are first connected in series and then in parallel. The ratio of heats produced in the two cases is

1

2 : 1

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

Assuming resistance of heater wire $= R$
Case I: When two heater wires are connected in series,
Heat produced $= H_{s} = V^{2} t / 2 R (1)$
Case II: When two heater wires are connected in parallel,
Net resistance $= R / 2$
$H_{p} = \frac{V^{2} t}{R / 2} (2)$
Dividing eqns. (1) and (2)
$\frac{H_{s}}{H_{p}} = \frac{V^{2} t}{2 R} \times \frac{R / 2}{V^{2} t} \Rightarrow \frac{H_{s}}{H_{p}} = \frac{1}{4}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357148 A $50 W$ bulb is in series with a room heater and the combination is connected across the mains. To get maximum heater output, the $50 W$ bulb should be replaced by

1

25 W

2

10 W

3

100 W

4

200 W

Explanation:

supporting img
For power across heater is maximum resistance of bulb should be minimum
$P_{heater} = (\frac{V}{R_{H} + R_{b}}) R_{H} R_{bulb}$ is minimum for $200 W$ .

PHXII03:CURRENT ELECTRICITY

357149 A current source drives a current in a coil of resistance $R_{1}$ for a time $t$ . The same source drives current in another coil of resistance $R_{2}$ for same time. If heat generated is same, find internal resistance of source.

1

\frac{R_{1} R_{2}}{R_{1} + R_{2}}

2

R_{1} + R_{2}

3

\sqrt{R_{1} R_{2}}

4

Z e r o

Explanation:

Let $r$ be the internal resistance of the source. The heat generated in a coil is
$H = P t = i^{2} R t$
Given that
$H_{1} = H_{2}$
$i_{1}^{2} R_{1} t = i_{2}^{2} R_{2} t$
${(\frac{ε}{R_{1} + r})}^{2} R_{1} = {(\frac{ε}{R_{2} + r})}^{2} R_{2}$
Where $ε$ is the emf of source.
$\Rightarrow r = \sqrt{R_{1} R_{2}}$

PHXII03:CURRENT ELECTRICITY

357146 In the circuit shown in figure, the heat produced in 5 $o h m$ resistance is 10 calories per second. The heat produced in $4 Ω$ resistance is
supporting img

1

3 c a l / \sec

2

4 c a l / \sec

3

1 c a l / \sec

4

2 c a l / \sec

Explanation:

Let $i_{1}$ and $i_{2}$ be the currents through the resistors.
$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{10}{5} = \frac{2}{1}$
Also heat produced per sec is, $P = i^{2} R$
$\Rightarrow \frac{P_{5}}{P_{4}} = {(\frac{i_{1}}{i_{2}})}^{2} \times \frac{5}{4} = {(\frac{2}{1})}^{2} \times \frac{5}{4} = \frac{5}{1}$
$\Rightarrow P_{4} = \frac{10}{5} = 2 c a l / \sec$
supporting img

PHXII03:CURRENT ELECTRICITY

357147 Two heater wires of equal lengths are first connected in series and then in parallel. The ratio of heats produced in the two cases is

1

2 : 1

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

Assuming resistance of heater wire $= R$
Case I: When two heater wires are connected in series,
Heat produced $= H_{s} = V^{2} t / 2 R (1)$
Case II: When two heater wires are connected in parallel,
Net resistance $= R / 2$
$H_{p} = \frac{V^{2} t}{R / 2} (2)$
Dividing eqns. (1) and (2)
$\frac{H_{s}}{H_{p}} = \frac{V^{2} t}{2 R} \times \frac{R / 2}{V^{2} t} \Rightarrow \frac{H_{s}}{H_{p}} = \frac{1}{4}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357148 A $50 W$ bulb is in series with a room heater and the combination is connected across the mains. To get maximum heater output, the $50 W$ bulb should be replaced by

1

25 W

2

10 W

3

100 W

4

200 W

Explanation:

supporting img
For power across heater is maximum resistance of bulb should be minimum
$P_{heater} = (\frac{V}{R_{H} + R_{b}}) R_{H} R_{bulb}$ is minimum for $200 W$ .

PHXII03:CURRENT ELECTRICITY

357149 A current source drives a current in a coil of resistance $R_{1}$ for a time $t$ . The same source drives current in another coil of resistance $R_{2}$ for same time. If heat generated is same, find internal resistance of source.

1

\frac{R_{1} R_{2}}{R_{1} + R_{2}}

2

R_{1} + R_{2}

3

\sqrt{R_{1} R_{2}}

4

Z e r o

Explanation:

Let $r$ be the internal resistance of the source. The heat generated in a coil is
$H = P t = i^{2} R t$
Given that
$H_{1} = H_{2}$
$i_{1}^{2} R_{1} t = i_{2}^{2} R_{2} t$
${(\frac{ε}{R_{1} + r})}^{2} R_{1} = {(\frac{ε}{R_{2} + r})}^{2} R_{2}$
Where $ε$ is the emf of source.
$\Rightarrow r = \sqrt{R_{1} R_{2}}$

PHXII03:CURRENT ELECTRICITY

357146 In the circuit shown in figure, the heat produced in 5 $o h m$ resistance is 10 calories per second. The heat produced in $4 Ω$ resistance is
supporting img

1

3 c a l / \sec

2

4 c a l / \sec

3

1 c a l / \sec

4

2 c a l / \sec

Explanation:

Let $i_{1}$ and $i_{2}$ be the currents through the resistors.
$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{10}{5} = \frac{2}{1}$
Also heat produced per sec is, $P = i^{2} R$
$\Rightarrow \frac{P_{5}}{P_{4}} = {(\frac{i_{1}}{i_{2}})}^{2} \times \frac{5}{4} = {(\frac{2}{1})}^{2} \times \frac{5}{4} = \frac{5}{1}$
$\Rightarrow P_{4} = \frac{10}{5} = 2 c a l / \sec$
supporting img

PHXII03:CURRENT ELECTRICITY

357147 Two heater wires of equal lengths are first connected in series and then in parallel. The ratio of heats produced in the two cases is

1

2 : 1

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

Assuming resistance of heater wire $= R$
Case I: When two heater wires are connected in series,
Heat produced $= H_{s} = V^{2} t / 2 R (1)$
Case II: When two heater wires are connected in parallel,
Net resistance $= R / 2$
$H_{p} = \frac{V^{2} t}{R / 2} (2)$
Dividing eqns. (1) and (2)
$\frac{H_{s}}{H_{p}} = \frac{V^{2} t}{2 R} \times \frac{R / 2}{V^{2} t} \Rightarrow \frac{H_{s}}{H_{p}} = \frac{1}{4}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357148 A $50 W$ bulb is in series with a room heater and the combination is connected across the mains. To get maximum heater output, the $50 W$ bulb should be replaced by

1

25 W

2

10 W

3

100 W

4

200 W

Explanation:

supporting img
For power across heater is maximum resistance of bulb should be minimum
$P_{heater} = (\frac{V}{R_{H} + R_{b}}) R_{H} R_{bulb}$ is minimum for $200 W$ .

PHXII03:CURRENT ELECTRICITY

357149 A current source drives a current in a coil of resistance $R_{1}$ for a time $t$ . The same source drives current in another coil of resistance $R_{2}$ for same time. If heat generated is same, find internal resistance of source.

1

\frac{R_{1} R_{2}}{R_{1} + R_{2}}

2

R_{1} + R_{2}

3

\sqrt{R_{1} R_{2}}

4

Z e r o

Explanation:

Let $r$ be the internal resistance of the source. The heat generated in a coil is
$H = P t = i^{2} R t$
Given that
$H_{1} = H_{2}$
$i_{1}^{2} R_{1} t = i_{2}^{2} R_{2} t$
${(\frac{ε}{R_{1} + r})}^{2} R_{1} = {(\frac{ε}{R_{2} + r})}^{2} R_{2}$
Where $ε$ is the emf of source.
$\Rightarrow r = \sqrt{R_{1} R_{2}}$

PHXII03:CURRENT ELECTRICITY

357146 In the circuit shown in figure, the heat produced in 5 $o h m$ resistance is 10 calories per second. The heat produced in $4 Ω$ resistance is
supporting img

1

3 c a l / \sec

2

4 c a l / \sec

3

1 c a l / \sec

4

2 c a l / \sec

Explanation:

Let $i_{1}$ and $i_{2}$ be the currents through the resistors.
$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{10}{5} = \frac{2}{1}$
Also heat produced per sec is, $P = i^{2} R$
$\Rightarrow \frac{P_{5}}{P_{4}} = {(\frac{i_{1}}{i_{2}})}^{2} \times \frac{5}{4} = {(\frac{2}{1})}^{2} \times \frac{5}{4} = \frac{5}{1}$
$\Rightarrow P_{4} = \frac{10}{5} = 2 c a l / \sec$
supporting img

PHXII03:CURRENT ELECTRICITY

357147 Two heater wires of equal lengths are first connected in series and then in parallel. The ratio of heats produced in the two cases is

1

2 : 1

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

Assuming resistance of heater wire $= R$
Case I: When two heater wires are connected in series,
Heat produced $= H_{s} = V^{2} t / 2 R (1)$
Case II: When two heater wires are connected in parallel,
Net resistance $= R / 2$
$H_{p} = \frac{V^{2} t}{R / 2} (2)$
Dividing eqns. (1) and (2)
$\frac{H_{s}}{H_{p}} = \frac{V^{2} t}{2 R} \times \frac{R / 2}{V^{2} t} \Rightarrow \frac{H_{s}}{H_{p}} = \frac{1}{4}$ .
So correct option is (2)

PHXII03:CURRENT ELECTRICITY

357148 A $50 W$ bulb is in series with a room heater and the combination is connected across the mains. To get maximum heater output, the $50 W$ bulb should be replaced by

1

25 W

2

10 W

3

100 W

4

200 W

Explanation:

supporting img
For power across heater is maximum resistance of bulb should be minimum
$P_{heater} = (\frac{V}{R_{H} + R_{b}}) R_{H} R_{bulb}$ is minimum for $200 W$ .

PHXII03:CURRENT ELECTRICITY

357149 A current source drives a current in a coil of resistance $R_{1}$ for a time $t$ . The same source drives current in another coil of resistance $R_{2}$ for same time. If heat generated is same, find internal resistance of source.

1

\frac{R_{1} R_{2}}{R_{1} + R_{2}}

2

R_{1} + R_{2}

3

\sqrt{R_{1} R_{2}}

4

Z e r o

Explanation:

Let $r$ be the internal resistance of the source. The heat generated in a coil is
$H = P t = i^{2} R t$
Given that
$H_{1} = H_{2}$
$i_{1}^{2} R_{1} t = i_{2}^{2} R_{2} t$
${(\frac{ε}{R_{1} + r})}^{2} R_{1} = {(\frac{ε}{R_{2} + r})}^{2} R_{2}$
Where $ε$ is the emf of source.
$\Rightarrow r = \sqrt{R_{1} R_{2}}$