QBManager

Current Electricity

152736 If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 W$ and $100 W$ electric bulbs operating at $220 V$ , then $(\frac{R_{1}}{R_{2}})$ is

1 1

2 2

3 0.5

4 4

5 0.25

Explanation:

C Given,
$P_{1} = 200 W, P_{2} = 100 W$
We know that,
$∵ Power, P = \frac{V^{2}}{R}$
$\frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100}{200} = \frac{1}{2} = 0.5$

Kerala CEE 2007

Current Electricity

152739 In the adjoining figure, if 10 calorie heat is produced per second in $5 Ω$ resister due to the flow of current through it, then the heat produced in $6 Ω$ resistor is

1

1 cal / s

2

2 cal / s

3

3 cal / s

4

4 cal / s

Explanation:

C We know that,

For $5 Ω$
$\frac{H}{t} = 10 cal / s$
Heat produced $(H) = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
$10 = I_{1}^{2} R$
$I_{1}^{2} = \frac{10}{5} = 2$
$I_{1} = \sqrt{2} A$
Voltage develop in $5 Ω$ resistance $=$ Voltage across $(6 +$ 4) $Ω$ registance
$I_{1} 5 = I_{2} (4 + 6)$
$I_{2} = \frac{5 I_{1}}{10} = \frac{5 \times \sqrt{2}}{10} = \frac{1}{\sqrt{2}} A$
$\frac{H}{t} = I_{2}^{2} R$ Heat produced in $6 Ω$ resistance
$\frac{H}{t} = {(\frac{1}{\sqrt{2}})}^{2} \times 6 = \frac{1}{2} \times 6 = 3 cal / s$

JCECE-2013

Current Electricity

152740 An electric heater boils $1 kg$ of water in a time $t_{1}$ . Another heater boils the same amount of water in a time $t_{2}$ . When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:

1

t_{1} + t_{2}

2

t_{1} t_{2}

3

\frac{t_{1} + t_{2}}{2}

4

\frac{t_{1} t_{2}}{t_{1} + t_{2}}

Explanation:

D We know that,
$R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 kg$ of water if $V$ is the voltage applied.
$H = \frac{V^{2}}{R_{1}} t_{1} = \frac{V^{2}}{R_{2}} t_{2}$
$\frac{t_{1}}{R_{1}} = \frac{t_{2}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{t_{1}}{t_{2}}$
When two heaters are in parallel and $t$ is time taken to boil the given water
$H = (\frac{V^{2}}{R_{1}} + \frac{V^{2}}{R_{2}}) t = \frac{V^{2}}{R_{1}} t_{1}$
$t = \frac{t_{1}}{R_{1}} {(\frac{1}{R_{1}} + \frac{1}{R_{2}})}^{- 1}$
$t = \frac{t_{1}}{1 + \frac{R_{1}}{R_{2}}}$
$t = \frac{t_{1}}{1 + \frac{t_{1}}{t_{2}}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

JCECE-2005]

Current Electricity

152741 An electric bulb is marked $100 W, 230 V$ . If the supply voltage drops to $115 V$ , what is the total energy produced by the bulb in $10 \min$ ?

1

30 kJ

2

20 kJ

3

15 kJ

4

10 kJ

Explanation:

C We have that,
Power of the bulb, $P = 100 W$
$V = 230 V$
$R$ be the resistance of the bulb
$P = \frac{V^{2}}{R} [∵ R = \frac{V^{2}}{P}]$
$R = \frac{230 \times 230}{100} = 529 Ω$
Changed supply voltage $V = 115 V$
Heat and light energy produced by the bulb in $10 \min$
$H = \frac{V^{2} t}{R} = \frac{115 \times 115 \times 10 \times 60}{529} = 15 kJ$

JCECE-2005

Current Electricity

152736 If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 W$ and $100 W$ electric bulbs operating at $220 V$ , then $(\frac{R_{1}}{R_{2}})$ is

1 1

2 2

3 0.5

4 4

5 0.25

Explanation:

C Given,
$P_{1} = 200 W, P_{2} = 100 W$
We know that,
$∵ Power, P = \frac{V^{2}}{R}$
$\frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100}{200} = \frac{1}{2} = 0.5$

Kerala CEE 2007

Current Electricity

152739 In the adjoining figure, if 10 calorie heat is produced per second in $5 Ω$ resister due to the flow of current through it, then the heat produced in $6 Ω$ resistor is

1

1 cal / s

2

2 cal / s

3

3 cal / s

4

4 cal / s

Explanation:

C We know that,

For $5 Ω$
$\frac{H}{t} = 10 cal / s$
Heat produced $(H) = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
$10 = I_{1}^{2} R$
$I_{1}^{2} = \frac{10}{5} = 2$
$I_{1} = \sqrt{2} A$
Voltage develop in $5 Ω$ resistance $=$ Voltage across $(6 +$ 4) $Ω$ registance
$I_{1} 5 = I_{2} (4 + 6)$
$I_{2} = \frac{5 I_{1}}{10} = \frac{5 \times \sqrt{2}}{10} = \frac{1}{\sqrt{2}} A$
$\frac{H}{t} = I_{2}^{2} R$ Heat produced in $6 Ω$ resistance
$\frac{H}{t} = {(\frac{1}{\sqrt{2}})}^{2} \times 6 = \frac{1}{2} \times 6 = 3 cal / s$

JCECE-2013

Current Electricity

152740 An electric heater boils $1 kg$ of water in a time $t_{1}$ . Another heater boils the same amount of water in a time $t_{2}$ . When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:

1

t_{1} + t_{2}

2

t_{1} t_{2}

3

\frac{t_{1} + t_{2}}{2}

4

\frac{t_{1} t_{2}}{t_{1} + t_{2}}

Explanation:

D We know that,
$R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 kg$ of water if $V$ is the voltage applied.
$H = \frac{V^{2}}{R_{1}} t_{1} = \frac{V^{2}}{R_{2}} t_{2}$
$\frac{t_{1}}{R_{1}} = \frac{t_{2}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{t_{1}}{t_{2}}$
When two heaters are in parallel and $t$ is time taken to boil the given water
$H = (\frac{V^{2}}{R_{1}} + \frac{V^{2}}{R_{2}}) t = \frac{V^{2}}{R_{1}} t_{1}$
$t = \frac{t_{1}}{R_{1}} {(\frac{1}{R_{1}} + \frac{1}{R_{2}})}^{- 1}$
$t = \frac{t_{1}}{1 + \frac{R_{1}}{R_{2}}}$
$t = \frac{t_{1}}{1 + \frac{t_{1}}{t_{2}}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

JCECE-2005]

Current Electricity

152741 An electric bulb is marked $100 W, 230 V$ . If the supply voltage drops to $115 V$ , what is the total energy produced by the bulb in $10 \min$ ?

1

30 kJ

2

20 kJ

3

15 kJ

4

10 kJ

Explanation:

C We have that,
Power of the bulb, $P = 100 W$
$V = 230 V$
$R$ be the resistance of the bulb
$P = \frac{V^{2}}{R} [∵ R = \frac{V^{2}}{P}]$
$R = \frac{230 \times 230}{100} = 529 Ω$
Changed supply voltage $V = 115 V$
Heat and light energy produced by the bulb in $10 \min$
$H = \frac{V^{2} t}{R} = \frac{115 \times 115 \times 10 \times 60}{529} = 15 kJ$

JCECE-2005

Current Electricity

152736 If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 W$ and $100 W$ electric bulbs operating at $220 V$ , then $(\frac{R_{1}}{R_{2}})$ is

1 1

2 2

3 0.5

4 4

5 0.25

Explanation:

C Given,
$P_{1} = 200 W, P_{2} = 100 W$
We know that,
$∵ Power, P = \frac{V^{2}}{R}$
$\frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100}{200} = \frac{1}{2} = 0.5$

Kerala CEE 2007

Current Electricity

152739 In the adjoining figure, if 10 calorie heat is produced per second in $5 Ω$ resister due to the flow of current through it, then the heat produced in $6 Ω$ resistor is

1

1 cal / s

2

2 cal / s

3

3 cal / s

4

4 cal / s

Explanation:

C We know that,

For $5 Ω$
$\frac{H}{t} = 10 cal / s$
Heat produced $(H) = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
$10 = I_{1}^{2} R$
$I_{1}^{2} = \frac{10}{5} = 2$
$I_{1} = \sqrt{2} A$
Voltage develop in $5 Ω$ resistance $=$ Voltage across $(6 +$ 4) $Ω$ registance
$I_{1} 5 = I_{2} (4 + 6)$
$I_{2} = \frac{5 I_{1}}{10} = \frac{5 \times \sqrt{2}}{10} = \frac{1}{\sqrt{2}} A$
$\frac{H}{t} = I_{2}^{2} R$ Heat produced in $6 Ω$ resistance
$\frac{H}{t} = {(\frac{1}{\sqrt{2}})}^{2} \times 6 = \frac{1}{2} \times 6 = 3 cal / s$

JCECE-2013

Current Electricity

152740 An electric heater boils $1 kg$ of water in a time $t_{1}$ . Another heater boils the same amount of water in a time $t_{2}$ . When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:

1

t_{1} + t_{2}

2

t_{1} t_{2}

3

\frac{t_{1} + t_{2}}{2}

4

\frac{t_{1} t_{2}}{t_{1} + t_{2}}

Explanation:

D We know that,
$R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 kg$ of water if $V$ is the voltage applied.
$H = \frac{V^{2}}{R_{1}} t_{1} = \frac{V^{2}}{R_{2}} t_{2}$
$\frac{t_{1}}{R_{1}} = \frac{t_{2}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{t_{1}}{t_{2}}$
When two heaters are in parallel and $t$ is time taken to boil the given water
$H = (\frac{V^{2}}{R_{1}} + \frac{V^{2}}{R_{2}}) t = \frac{V^{2}}{R_{1}} t_{1}$
$t = \frac{t_{1}}{R_{1}} {(\frac{1}{R_{1}} + \frac{1}{R_{2}})}^{- 1}$
$t = \frac{t_{1}}{1 + \frac{R_{1}}{R_{2}}}$
$t = \frac{t_{1}}{1 + \frac{t_{1}}{t_{2}}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

JCECE-2005]

Current Electricity

152741 An electric bulb is marked $100 W, 230 V$ . If the supply voltage drops to $115 V$ , what is the total energy produced by the bulb in $10 \min$ ?

1

30 kJ

2

20 kJ

3

15 kJ

4

10 kJ

Explanation:

C We have that,
Power of the bulb, $P = 100 W$
$V = 230 V$
$R$ be the resistance of the bulb
$P = \frac{V^{2}}{R} [∵ R = \frac{V^{2}}{P}]$
$R = \frac{230 \times 230}{100} = 529 Ω$
Changed supply voltage $V = 115 V$
Heat and light energy produced by the bulb in $10 \min$
$H = \frac{V^{2} t}{R} = \frac{115 \times 115 \times 10 \times 60}{529} = 15 kJ$

JCECE-2005

Current Electricity

152736 If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 W$ and $100 W$ electric bulbs operating at $220 V$ , then $(\frac{R_{1}}{R_{2}})$ is

1 1

2 2

3 0.5

4 4

5 0.25

Explanation:

C Given,
$P_{1} = 200 W, P_{2} = 100 W$
We know that,
$∵ Power, P = \frac{V^{2}}{R}$
$\frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100}{200} = \frac{1}{2} = 0.5$

Kerala CEE 2007

Current Electricity

152739 In the adjoining figure, if 10 calorie heat is produced per second in $5 Ω$ resister due to the flow of current through it, then the heat produced in $6 Ω$ resistor is

1

1 cal / s

2

2 cal / s

3

3 cal / s

4

4 cal / s

Explanation:

C We know that,

For $5 Ω$
$\frac{H}{t} = 10 cal / s$
Heat produced $(H) = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
$10 = I_{1}^{2} R$
$I_{1}^{2} = \frac{10}{5} = 2$
$I_{1} = \sqrt{2} A$
Voltage develop in $5 Ω$ resistance $=$ Voltage across $(6 +$ 4) $Ω$ registance
$I_{1} 5 = I_{2} (4 + 6)$
$I_{2} = \frac{5 I_{1}}{10} = \frac{5 \times \sqrt{2}}{10} = \frac{1}{\sqrt{2}} A$
$\frac{H}{t} = I_{2}^{2} R$ Heat produced in $6 Ω$ resistance
$\frac{H}{t} = {(\frac{1}{\sqrt{2}})}^{2} \times 6 = \frac{1}{2} \times 6 = 3 cal / s$

JCECE-2013

Current Electricity

152740 An electric heater boils $1 kg$ of water in a time $t_{1}$ . Another heater boils the same amount of water in a time $t_{2}$ . When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:

1

t_{1} + t_{2}

2

t_{1} t_{2}

3

\frac{t_{1} + t_{2}}{2}

4

\frac{t_{1} t_{2}}{t_{1} + t_{2}}

Explanation:

D We know that,
$R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 kg$ of water if $V$ is the voltage applied.
$H = \frac{V^{2}}{R_{1}} t_{1} = \frac{V^{2}}{R_{2}} t_{2}$
$\frac{t_{1}}{R_{1}} = \frac{t_{2}}{R_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{t_{1}}{t_{2}}$
When two heaters are in parallel and $t$ is time taken to boil the given water
$H = (\frac{V^{2}}{R_{1}} + \frac{V^{2}}{R_{2}}) t = \frac{V^{2}}{R_{1}} t_{1}$
$t = \frac{t_{1}}{R_{1}} {(\frac{1}{R_{1}} + \frac{1}{R_{2}})}^{- 1}$
$t = \frac{t_{1}}{1 + \frac{R_{1}}{R_{2}}}$
$t = \frac{t_{1}}{1 + \frac{t_{1}}{t_{2}}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

JCECE-2005]

Current Electricity

152741 An electric bulb is marked $100 W, 230 V$ . If the supply voltage drops to $115 V$ , what is the total energy produced by the bulb in $10 \min$ ?

1

30 kJ

2

20 kJ

3

15 kJ

4

10 kJ

Explanation:

C We have that,
Power of the bulb, $P = 100 W$
$V = 230 V$
$R$ be the resistance of the bulb
$P = \frac{V^{2}}{R} [∵ R = \frac{V^{2}}{P}]$
$R = \frac{230 \times 230}{100} = 529 Ω$
Changed supply voltage $V = 115 V$
Heat and light energy produced by the bulb in $10 \min$
$H = \frac{V^{2} t}{R} = \frac{115 \times 115 \times 10 \times 60}{529} = 15 kJ$

JCECE-2005

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