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Current Electricity

152640 Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

A When heater filament are connected in parallel,
$H_{P} = \frac{V^{2}}{R_{P}} t$
$H_{P} = \frac{V^{2}}{\frac{R}{2} t}$
$H_{P} = \frac{2 V^{2}}{R} t$
When heater filament are connected in series,
$H_{S} = \frac{V^{2}}{R_{S}} t$
$H_{S} = \frac{V^{2}}{2 R} t$
On dividing equation (i) by (ii), we get-
$\frac{H_{P}}{H_{S}} = \frac{2 V^{2} t \times 2 R}{R \times V^{2} \times t}$
$\frac{H_{P}}{H_{S}} = \frac{4}{1} or 4 : 1$

JEE Main-11.04.2023

Current Electricity

152641 Ratio of thermal energy released in two resistors $R$ and $3 R$ connected in parallel in an electric circuit is:

1

1 : 3

2

3 : 1

3

1 : 1

4

1 : 27

Explanation:

B Since, the potential difference are same and resistance are in parallel.
$∵ H = \frac{V^{2}}{R} \times t$
$H_{1} = \frac{V^{2}}{R} t & H_{2} = \frac{V^{2}}{3 R} t$
Now, the ratio of thermal energy-
$\frac{H_{1}}{H_{2}} = \frac{\frac{V^{2} t}{R}}{\frac{V^{2} t}{3 R}} = \frac{V^{2} t \times 3 R}{V^{2} t \times R}$
$\frac{H_{1}}{H_{2}} = \frac{3}{1}$
$H_{1} : H_{2} = 3 : 1$

JEE Main-29.01.2023

Current Electricity

152642 Currents flowing through two conductors of equal resistance produce 9 times more heat in first conductor compared to the second one. The ratio of currents flowing through the first and second conductors is

1

1 : 3

2

3 : 1

3

1 : 9

4

9 : 1

Explanation:

A Given,
Heat produce in $2^{nd}$ conductor $= 9 \times$ Heat produce in $1^{st}$ conductor
$Q_{2} = 9 Q_{1}$
We know that,
Heat produced through conductor
$(Q) = I^{2} R$
$∴ Q \propto I^{2}$
Now, the ratio of current flowing $\frac{I_{1}}{I_{2}} = \sqrt{\frac{Q_{1}}{Q_{2}}} = \sqrt{\frac{Q_{1}}{9 Q_{1}}}$
$\frac{I_{1}}{I_{2}} = \frac{1}{3}$

AP EAMCET-11.07.2022

Current Electricity

152643 An electric bulb is connected to a $220 V$ generator. The current drawn is $600 mA$ . What is the power of the bulb?

1

132 W

2

13 \cdot 2 W

3

1320 W

4

13200 W

Explanation:

A Given,
Voltage $(V) = 220$ Volt
Current (i) $= 600 mA = 0.6 A$
We know that,
Power $(P) =$ Voltage $\times$ Current
$P = 220 \times 0.6$
$P = 132 W$

UPSC NDA-04.09.2022

Current Electricity

152644 A heater of $9 Ω$ resistance is connected to $30 V$ DC supply for 1 minute. Energy produced by the heater is

1

100 J

2

270 J

3

600 J

4

6000 J

Explanation:

D Given that,
$R = 9 Ω, V = 30 V, T = 60 \sec$
We know that, Ohm's law -
$V = IR$
$I = \frac{V}{R}$
$I = \frac{30}{9} = \frac{10}{3} A .$
Using the formula,
$H = I^{2} R T$
$H = \frac{10}{3} \times \frac{10}{3} \times 9 \times 60$
$H = 100 \times 60 = 6000 J$

Tripura-27.04.2022

Current Electricity

152640 Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

A When heater filament are connected in parallel,
$H_{P} = \frac{V^{2}}{R_{P}} t$
$H_{P} = \frac{V^{2}}{\frac{R}{2} t}$
$H_{P} = \frac{2 V^{2}}{R} t$
When heater filament are connected in series,
$H_{S} = \frac{V^{2}}{R_{S}} t$
$H_{S} = \frac{V^{2}}{2 R} t$
On dividing equation (i) by (ii), we get-
$\frac{H_{P}}{H_{S}} = \frac{2 V^{2} t \times 2 R}{R \times V^{2} \times t}$
$\frac{H_{P}}{H_{S}} = \frac{4}{1} or 4 : 1$

JEE Main-11.04.2023

Current Electricity

152641 Ratio of thermal energy released in two resistors $R$ and $3 R$ connected in parallel in an electric circuit is:

1

1 : 3

2

3 : 1

3

1 : 1

4

1 : 27

Explanation:

B Since, the potential difference are same and resistance are in parallel.
$∵ H = \frac{V^{2}}{R} \times t$
$H_{1} = \frac{V^{2}}{R} t & H_{2} = \frac{V^{2}}{3 R} t$
Now, the ratio of thermal energy-
$\frac{H_{1}}{H_{2}} = \frac{\frac{V^{2} t}{R}}{\frac{V^{2} t}{3 R}} = \frac{V^{2} t \times 3 R}{V^{2} t \times R}$
$\frac{H_{1}}{H_{2}} = \frac{3}{1}$
$H_{1} : H_{2} = 3 : 1$

JEE Main-29.01.2023

Current Electricity

152642 Currents flowing through two conductors of equal resistance produce 9 times more heat in first conductor compared to the second one. The ratio of currents flowing through the first and second conductors is

1

1 : 3

2

3 : 1

3

1 : 9

4

9 : 1

Explanation:

A Given,
Heat produce in $2^{nd}$ conductor $= 9 \times$ Heat produce in $1^{st}$ conductor
$Q_{2} = 9 Q_{1}$
We know that,
Heat produced through conductor
$(Q) = I^{2} R$
$∴ Q \propto I^{2}$
Now, the ratio of current flowing $\frac{I_{1}}{I_{2}} = \sqrt{\frac{Q_{1}}{Q_{2}}} = \sqrt{\frac{Q_{1}}{9 Q_{1}}}$
$\frac{I_{1}}{I_{2}} = \frac{1}{3}$

AP EAMCET-11.07.2022

Current Electricity

152643 An electric bulb is connected to a $220 V$ generator. The current drawn is $600 mA$ . What is the power of the bulb?

1

132 W

2

13 \cdot 2 W

3

1320 W

4

13200 W

Explanation:

A Given,
Voltage $(V) = 220$ Volt
Current (i) $= 600 mA = 0.6 A$
We know that,
Power $(P) =$ Voltage $\times$ Current
$P = 220 \times 0.6$
$P = 132 W$

UPSC NDA-04.09.2022

Current Electricity

152644 A heater of $9 Ω$ resistance is connected to $30 V$ DC supply for 1 minute. Energy produced by the heater is

1

100 J

2

270 J

3

600 J

4

6000 J

Explanation:

D Given that,
$R = 9 Ω, V = 30 V, T = 60 \sec$
We know that, Ohm's law -
$V = IR$
$I = \frac{V}{R}$
$I = \frac{30}{9} = \frac{10}{3} A .$
Using the formula,
$H = I^{2} R T$
$H = \frac{10}{3} \times \frac{10}{3} \times 9 \times 60$
$H = 100 \times 60 = 6000 J$

Tripura-27.04.2022

Current Electricity

152640 Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

A When heater filament are connected in parallel,
$H_{P} = \frac{V^{2}}{R_{P}} t$
$H_{P} = \frac{V^{2}}{\frac{R}{2} t}$
$H_{P} = \frac{2 V^{2}}{R} t$
When heater filament are connected in series,
$H_{S} = \frac{V^{2}}{R_{S}} t$
$H_{S} = \frac{V^{2}}{2 R} t$
On dividing equation (i) by (ii), we get-
$\frac{H_{P}}{H_{S}} = \frac{2 V^{2} t \times 2 R}{R \times V^{2} \times t}$
$\frac{H_{P}}{H_{S}} = \frac{4}{1} or 4 : 1$

JEE Main-11.04.2023

Current Electricity

152641 Ratio of thermal energy released in two resistors $R$ and $3 R$ connected in parallel in an electric circuit is:

1

1 : 3

2

3 : 1

3

1 : 1

4

1 : 27

Explanation:

B Since, the potential difference are same and resistance are in parallel.
$∵ H = \frac{V^{2}}{R} \times t$
$H_{1} = \frac{V^{2}}{R} t & H_{2} = \frac{V^{2}}{3 R} t$
Now, the ratio of thermal energy-
$\frac{H_{1}}{H_{2}} = \frac{\frac{V^{2} t}{R}}{\frac{V^{2} t}{3 R}} = \frac{V^{2} t \times 3 R}{V^{2} t \times R}$
$\frac{H_{1}}{H_{2}} = \frac{3}{1}$
$H_{1} : H_{2} = 3 : 1$

JEE Main-29.01.2023

Current Electricity

152642 Currents flowing through two conductors of equal resistance produce 9 times more heat in first conductor compared to the second one. The ratio of currents flowing through the first and second conductors is

1

1 : 3

2

3 : 1

3

1 : 9

4

9 : 1

Explanation:

A Given,
Heat produce in $2^{nd}$ conductor $= 9 \times$ Heat produce in $1^{st}$ conductor
$Q_{2} = 9 Q_{1}$
We know that,
Heat produced through conductor
$(Q) = I^{2} R$
$∴ Q \propto I^{2}$
Now, the ratio of current flowing $\frac{I_{1}}{I_{2}} = \sqrt{\frac{Q_{1}}{Q_{2}}} = \sqrt{\frac{Q_{1}}{9 Q_{1}}}$
$\frac{I_{1}}{I_{2}} = \frac{1}{3}$

AP EAMCET-11.07.2022

Current Electricity

152643 An electric bulb is connected to a $220 V$ generator. The current drawn is $600 mA$ . What is the power of the bulb?

1

132 W

2

13 \cdot 2 W

3

1320 W

4

13200 W

Explanation:

A Given,
Voltage $(V) = 220$ Volt
Current (i) $= 600 mA = 0.6 A$
We know that,
Power $(P) =$ Voltage $\times$ Current
$P = 220 \times 0.6$
$P = 132 W$

UPSC NDA-04.09.2022

Current Electricity

152644 A heater of $9 Ω$ resistance is connected to $30 V$ DC supply for 1 minute. Energy produced by the heater is

1

100 J

2

270 J

3

600 J

4

6000 J

Explanation:

D Given that,
$R = 9 Ω, V = 30 V, T = 60 \sec$
We know that, Ohm's law -
$V = IR$
$I = \frac{V}{R}$
$I = \frac{30}{9} = \frac{10}{3} A .$
Using the formula,
$H = I^{2} R T$
$H = \frac{10}{3} \times \frac{10}{3} \times 9 \times 60$
$H = 100 \times 60 = 6000 J$

Tripura-27.04.2022

Current Electricity

152640 Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

A When heater filament are connected in parallel,
$H_{P} = \frac{V^{2}}{R_{P}} t$
$H_{P} = \frac{V^{2}}{\frac{R}{2} t}$
$H_{P} = \frac{2 V^{2}}{R} t$
When heater filament are connected in series,
$H_{S} = \frac{V^{2}}{R_{S}} t$
$H_{S} = \frac{V^{2}}{2 R} t$
On dividing equation (i) by (ii), we get-
$\frac{H_{P}}{H_{S}} = \frac{2 V^{2} t \times 2 R}{R \times V^{2} \times t}$
$\frac{H_{P}}{H_{S}} = \frac{4}{1} or 4 : 1$

JEE Main-11.04.2023

Current Electricity

152641 Ratio of thermal energy released in two resistors $R$ and $3 R$ connected in parallel in an electric circuit is:

1

1 : 3

2

3 : 1

3

1 : 1

4

1 : 27

Explanation:

B Since, the potential difference are same and resistance are in parallel.
$∵ H = \frac{V^{2}}{R} \times t$
$H_{1} = \frac{V^{2}}{R} t & H_{2} = \frac{V^{2}}{3 R} t$
Now, the ratio of thermal energy-
$\frac{H_{1}}{H_{2}} = \frac{\frac{V^{2} t}{R}}{\frac{V^{2} t}{3 R}} = \frac{V^{2} t \times 3 R}{V^{2} t \times R}$
$\frac{H_{1}}{H_{2}} = \frac{3}{1}$
$H_{1} : H_{2} = 3 : 1$

JEE Main-29.01.2023

Current Electricity

152642 Currents flowing through two conductors of equal resistance produce 9 times more heat in first conductor compared to the second one. The ratio of currents flowing through the first and second conductors is

1

1 : 3

2

3 : 1

3

1 : 9

4

9 : 1

Explanation:

A Given,
Heat produce in $2^{nd}$ conductor $= 9 \times$ Heat produce in $1^{st}$ conductor
$Q_{2} = 9 Q_{1}$
We know that,
Heat produced through conductor
$(Q) = I^{2} R$
$∴ Q \propto I^{2}$
Now, the ratio of current flowing $\frac{I_{1}}{I_{2}} = \sqrt{\frac{Q_{1}}{Q_{2}}} = \sqrt{\frac{Q_{1}}{9 Q_{1}}}$
$\frac{I_{1}}{I_{2}} = \frac{1}{3}$

AP EAMCET-11.07.2022

Current Electricity

152643 An electric bulb is connected to a $220 V$ generator. The current drawn is $600 mA$ . What is the power of the bulb?

1

132 W

2

13 \cdot 2 W

3

1320 W

4

13200 W

Explanation:

A Given,
Voltage $(V) = 220$ Volt
Current (i) $= 600 mA = 0.6 A$
We know that,
Power $(P) =$ Voltage $\times$ Current
$P = 220 \times 0.6$
$P = 132 W$

UPSC NDA-04.09.2022

Current Electricity

152644 A heater of $9 Ω$ resistance is connected to $30 V$ DC supply for 1 minute. Energy produced by the heater is

1

100 J

2

270 J

3

600 J

4

6000 J

Explanation:

D Given that,
$R = 9 Ω, V = 30 V, T = 60 \sec$
We know that, Ohm's law -
$V = IR$
$I = \frac{V}{R}$
$I = \frac{30}{9} = \frac{10}{3} A .$
Using the formula,
$H = I^{2} R T$
$H = \frac{10}{3} \times \frac{10}{3} \times 9 \times 60$
$H = 100 \times 60 = 6000 J$

Tripura-27.04.2022

Current Electricity

152640 Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

A When heater filament are connected in parallel,
$H_{P} = \frac{V^{2}}{R_{P}} t$
$H_{P} = \frac{V^{2}}{\frac{R}{2} t}$
$H_{P} = \frac{2 V^{2}}{R} t$
When heater filament are connected in series,
$H_{S} = \frac{V^{2}}{R_{S}} t$
$H_{S} = \frac{V^{2}}{2 R} t$
On dividing equation (i) by (ii), we get-
$\frac{H_{P}}{H_{S}} = \frac{2 V^{2} t \times 2 R}{R \times V^{2} \times t}$
$\frac{H_{P}}{H_{S}} = \frac{4}{1} or 4 : 1$

JEE Main-11.04.2023

Current Electricity

152641 Ratio of thermal energy released in two resistors $R$ and $3 R$ connected in parallel in an electric circuit is:

1

1 : 3

2

3 : 1

3

1 : 1

4

1 : 27

Explanation:

B Since, the potential difference are same and resistance are in parallel.
$∵ H = \frac{V^{2}}{R} \times t$
$H_{1} = \frac{V^{2}}{R} t & H_{2} = \frac{V^{2}}{3 R} t$
Now, the ratio of thermal energy-
$\frac{H_{1}}{H_{2}} = \frac{\frac{V^{2} t}{R}}{\frac{V^{2} t}{3 R}} = \frac{V^{2} t \times 3 R}{V^{2} t \times R}$
$\frac{H_{1}}{H_{2}} = \frac{3}{1}$
$H_{1} : H_{2} = 3 : 1$

JEE Main-29.01.2023

Current Electricity

152642 Currents flowing through two conductors of equal resistance produce 9 times more heat in first conductor compared to the second one. The ratio of currents flowing through the first and second conductors is

1

1 : 3

2

3 : 1

3

1 : 9

4

9 : 1

Explanation:

A Given,
Heat produce in $2^{nd}$ conductor $= 9 \times$ Heat produce in $1^{st}$ conductor
$Q_{2} = 9 Q_{1}$
We know that,
Heat produced through conductor
$(Q) = I^{2} R$
$∴ Q \propto I^{2}$
Now, the ratio of current flowing $\frac{I_{1}}{I_{2}} = \sqrt{\frac{Q_{1}}{Q_{2}}} = \sqrt{\frac{Q_{1}}{9 Q_{1}}}$
$\frac{I_{1}}{I_{2}} = \frac{1}{3}$

AP EAMCET-11.07.2022

Current Electricity

152643 An electric bulb is connected to a $220 V$ generator. The current drawn is $600 mA$ . What is the power of the bulb?

1

132 W

2

13 \cdot 2 W

3

1320 W

4

13200 W

Explanation:

A Given,
Voltage $(V) = 220$ Volt
Current (i) $= 600 mA = 0.6 A$
We know that,
Power $(P) =$ Voltage $\times$ Current
$P = 220 \times 0.6$
$P = 132 W$

UPSC NDA-04.09.2022

Current Electricity

152644 A heater of $9 Ω$ resistance is connected to $30 V$ DC supply for 1 minute. Energy produced by the heater is

1

100 J

2

270 J

3

600 J

4

6000 J

Explanation:

D Given that,
$R = 9 Ω, V = 30 V, T = 60 \sec$
We know that, Ohm's law -
$V = IR$
$I = \frac{V}{R}$
$I = \frac{30}{9} = \frac{10}{3} A .$
Using the formula,
$H = I^{2} R T$
$H = \frac{10}{3} \times \frac{10}{3} \times 9 \times 60$
$H = 100 \times 60 = 6000 J$

Tripura-27.04.2022