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Current Electricity

152747 Assertion: Long distance power transmission is done at high voltage.
Reason: At high voltage supply power losses are less.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A As to transmit power over long distance the transmission is done high voltages supply power losses are less.
$P_{loss} = I^{2} R = {(\frac{P}{V})}^{2} R$
Power loss is inversely proportional to the square of the voltage.

AIIMS-2014

Current Electricity

152748 Assertion: When current through a bulb decreases by $0.5 %$ , the glow of bulb decreases by $1 %$ .
Reason: Glow (Power) which is directly proportional to square of current.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A We know that,
Power, $P = I^{2} R$
$\frac{dP}{P} = 2 \frac{dI}{I}$
$= 2 \times 0.5 % = 1 %$
Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 %$ when current through the bulb decreases by $0.5 %$ .

AIIMS-2015

Current Electricity

152750 A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if

1 both the length and radius of wire are halved

2 both length and radius of wire are doubled

3 the radius of wire is doubled

4 the length of the wire is doubled.

Explanation:

B The heat produced is given by,
$H = \frac{V^{2}}{R} t$
We know that,
$R = \frac{ρ l}{π r^{2}}$
Where, Length of wire $= l$
$Radius of wire = r$
$ρ = resistivity$
On putting the value $R$ in equation (i), We get -
$Then, H = \frac{V^{2}}{\frac{ρ l}{π r^{2}}} t$
$H = V^{2} (\frac{π r^{2}}{ρ l}) t$
When, $l$ and $r$ are doubled.
Then, $H^{'} = \frac{V^{2} π}{ρ} (\frac{(2 r)^{2}}{2 l}) t$
$H^{'} = \frac{V^{2}}{ρ} (\frac{π r^{2}}{l}) t \times 2$
$H^{'} = 2 H$
So, heat produced will be doubled when both the length and radius of the wire are doubled.

AIIMS-2012

Current Electricity

152749 Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on.
Reason: Dimness decreases after sometime.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

AIIMS-2008

Current Electricity

152751 In the circuit shown in figure, the $5 Ω$ resistance develops $20.00 cal / s$ due to the current flowing through it. The heat developed in $2 Ω$ resistance (in $cal / s$ ) is

1 23.8

2 14.2

3 11.9

4 7.1

Explanation:

B Given, original image
$Heat, H = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
For $R = 5 Ω$ , and $\frac{H}{t} = 20.00 cal / s$
$∴ 20 = I_{1}^{2} \times 5$
$I_{1} = 2 A$
Voltage across $5 Ω =$ Voltage across $(6 + 9) Ω$
Now, $V = I_{1} \times 5 = I_{2} \times (6 + 9)$
$5 I_{1} = 15 I_{2}$
$∴ I_{2} = \frac{5 I_{1}}{15}$
$I_{2} = \frac{5 \times 2}{15}$
$I_{2} = \frac{2}{3} A$
Hence, heat in $2 Ω$ resistor per second,
$\frac{H}{t} = {(I_{1} + I_{2})}^{2} \times 2$
$\frac{H}{t} = {(2 + \frac{2}{3})}^{2} \times 2$
$\frac{H}{t} = {(\frac{8}{3})}^{2} \times 2$
$\frac{H}{t} = 14.2 cal / s$

AIIMS-2012

Current Electricity

152747 Assertion: Long distance power transmission is done at high voltage.
Reason: At high voltage supply power losses are less.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A As to transmit power over long distance the transmission is done high voltages supply power losses are less.
$P_{loss} = I^{2} R = {(\frac{P}{V})}^{2} R$
Power loss is inversely proportional to the square of the voltage.

AIIMS-2014

Current Electricity

152748 Assertion: When current through a bulb decreases by $0.5 %$ , the glow of bulb decreases by $1 %$ .
Reason: Glow (Power) which is directly proportional to square of current.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A We know that,
Power, $P = I^{2} R$
$\frac{dP}{P} = 2 \frac{dI}{I}$
$= 2 \times 0.5 % = 1 %$
Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 %$ when current through the bulb decreases by $0.5 %$ .

AIIMS-2015

Current Electricity

152750 A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if

1 both the length and radius of wire are halved

2 both length and radius of wire are doubled

3 the radius of wire is doubled

4 the length of the wire is doubled.

Explanation:

B The heat produced is given by,
$H = \frac{V^{2}}{R} t$
We know that,
$R = \frac{ρ l}{π r^{2}}$
Where, Length of wire $= l$
$Radius of wire = r$
$ρ = resistivity$
On putting the value $R$ in equation (i), We get -
$Then, H = \frac{V^{2}}{\frac{ρ l}{π r^{2}}} t$
$H = V^{2} (\frac{π r^{2}}{ρ l}) t$
When, $l$ and $r$ are doubled.
Then, $H^{'} = \frac{V^{2} π}{ρ} (\frac{(2 r)^{2}}{2 l}) t$
$H^{'} = \frac{V^{2}}{ρ} (\frac{π r^{2}}{l}) t \times 2$
$H^{'} = 2 H$
So, heat produced will be doubled when both the length and radius of the wire are doubled.

AIIMS-2012

Current Electricity

152749 Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on.
Reason: Dimness decreases after sometime.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

AIIMS-2008

Current Electricity

152751 In the circuit shown in figure, the $5 Ω$ resistance develops $20.00 cal / s$ due to the current flowing through it. The heat developed in $2 Ω$ resistance (in $cal / s$ ) is

1 23.8

2 14.2

3 11.9

4 7.1

Explanation:

B Given, original image
$Heat, H = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
For $R = 5 Ω$ , and $\frac{H}{t} = 20.00 cal / s$
$∴ 20 = I_{1}^{2} \times 5$
$I_{1} = 2 A$
Voltage across $5 Ω =$ Voltage across $(6 + 9) Ω$
Now, $V = I_{1} \times 5 = I_{2} \times (6 + 9)$
$5 I_{1} = 15 I_{2}$
$∴ I_{2} = \frac{5 I_{1}}{15}$
$I_{2} = \frac{5 \times 2}{15}$
$I_{2} = \frac{2}{3} A$
Hence, heat in $2 Ω$ resistor per second,
$\frac{H}{t} = {(I_{1} + I_{2})}^{2} \times 2$
$\frac{H}{t} = {(2 + \frac{2}{3})}^{2} \times 2$
$\frac{H}{t} = {(\frac{8}{3})}^{2} \times 2$
$\frac{H}{t} = 14.2 cal / s$

AIIMS-2012

Current Electricity

152747 Assertion: Long distance power transmission is done at high voltage.
Reason: At high voltage supply power losses are less.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A As to transmit power over long distance the transmission is done high voltages supply power losses are less.
$P_{loss} = I^{2} R = {(\frac{P}{V})}^{2} R$
Power loss is inversely proportional to the square of the voltage.

AIIMS-2014

Current Electricity

152748 Assertion: When current through a bulb decreases by $0.5 %$ , the glow of bulb decreases by $1 %$ .
Reason: Glow (Power) which is directly proportional to square of current.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A We know that,
Power, $P = I^{2} R$
$\frac{dP}{P} = 2 \frac{dI}{I}$
$= 2 \times 0.5 % = 1 %$
Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 %$ when current through the bulb decreases by $0.5 %$ .

AIIMS-2015

Current Electricity

152750 A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if

1 both the length and radius of wire are halved

2 both length and radius of wire are doubled

3 the radius of wire is doubled

4 the length of the wire is doubled.

Explanation:

B The heat produced is given by,
$H = \frac{V^{2}}{R} t$
We know that,
$R = \frac{ρ l}{π r^{2}}$
Where, Length of wire $= l$
$Radius of wire = r$
$ρ = resistivity$
On putting the value $R$ in equation (i), We get -
$Then, H = \frac{V^{2}}{\frac{ρ l}{π r^{2}}} t$
$H = V^{2} (\frac{π r^{2}}{ρ l}) t$
When, $l$ and $r$ are doubled.
Then, $H^{'} = \frac{V^{2} π}{ρ} (\frac{(2 r)^{2}}{2 l}) t$
$H^{'} = \frac{V^{2}}{ρ} (\frac{π r^{2}}{l}) t \times 2$
$H^{'} = 2 H$
So, heat produced will be doubled when both the length and radius of the wire are doubled.

AIIMS-2012

Current Electricity

152749 Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on.
Reason: Dimness decreases after sometime.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

AIIMS-2008

Current Electricity

152751 In the circuit shown in figure, the $5 Ω$ resistance develops $20.00 cal / s$ due to the current flowing through it. The heat developed in $2 Ω$ resistance (in $cal / s$ ) is

1 23.8

2 14.2

3 11.9

4 7.1

Explanation:

B Given, original image
$Heat, H = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
For $R = 5 Ω$ , and $\frac{H}{t} = 20.00 cal / s$
$∴ 20 = I_{1}^{2} \times 5$
$I_{1} = 2 A$
Voltage across $5 Ω =$ Voltage across $(6 + 9) Ω$
Now, $V = I_{1} \times 5 = I_{2} \times (6 + 9)$
$5 I_{1} = 15 I_{2}$
$∴ I_{2} = \frac{5 I_{1}}{15}$
$I_{2} = \frac{5 \times 2}{15}$
$I_{2} = \frac{2}{3} A$
Hence, heat in $2 Ω$ resistor per second,
$\frac{H}{t} = {(I_{1} + I_{2})}^{2} \times 2$
$\frac{H}{t} = {(2 + \frac{2}{3})}^{2} \times 2$
$\frac{H}{t} = {(\frac{8}{3})}^{2} \times 2$
$\frac{H}{t} = 14.2 cal / s$

AIIMS-2012

Current Electricity

152747 Assertion: Long distance power transmission is done at high voltage.
Reason: At high voltage supply power losses are less.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A As to transmit power over long distance the transmission is done high voltages supply power losses are less.
$P_{loss} = I^{2} R = {(\frac{P}{V})}^{2} R$
Power loss is inversely proportional to the square of the voltage.

AIIMS-2014

Current Electricity

152748 Assertion: When current through a bulb decreases by $0.5 %$ , the glow of bulb decreases by $1 %$ .
Reason: Glow (Power) which is directly proportional to square of current.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A We know that,
Power, $P = I^{2} R$
$\frac{dP}{P} = 2 \frac{dI}{I}$
$= 2 \times 0.5 % = 1 %$
Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 %$ when current through the bulb decreases by $0.5 %$ .

AIIMS-2015

Current Electricity

152750 A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if

1 both the length and radius of wire are halved

2 both length and radius of wire are doubled

3 the radius of wire is doubled

4 the length of the wire is doubled.

Explanation:

B The heat produced is given by,
$H = \frac{V^{2}}{R} t$
We know that,
$R = \frac{ρ l}{π r^{2}}$
Where, Length of wire $= l$
$Radius of wire = r$
$ρ = resistivity$
On putting the value $R$ in equation (i), We get -
$Then, H = \frac{V^{2}}{\frac{ρ l}{π r^{2}}} t$
$H = V^{2} (\frac{π r^{2}}{ρ l}) t$
When, $l$ and $r$ are doubled.
Then, $H^{'} = \frac{V^{2} π}{ρ} (\frac{(2 r)^{2}}{2 l}) t$
$H^{'} = \frac{V^{2}}{ρ} (\frac{π r^{2}}{l}) t \times 2$
$H^{'} = 2 H$
So, heat produced will be doubled when both the length and radius of the wire are doubled.

AIIMS-2012

Current Electricity

152749 Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on.
Reason: Dimness decreases after sometime.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

AIIMS-2008

Current Electricity

152751 In the circuit shown in figure, the $5 Ω$ resistance develops $20.00 cal / s$ due to the current flowing through it. The heat developed in $2 Ω$ resistance (in $cal / s$ ) is

1 23.8

2 14.2

3 11.9

4 7.1

Explanation:

B Given, original image
$Heat, H = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
For $R = 5 Ω$ , and $\frac{H}{t} = 20.00 cal / s$
$∴ 20 = I_{1}^{2} \times 5$
$I_{1} = 2 A$
Voltage across $5 Ω =$ Voltage across $(6 + 9) Ω$
Now, $V = I_{1} \times 5 = I_{2} \times (6 + 9)$
$5 I_{1} = 15 I_{2}$
$∴ I_{2} = \frac{5 I_{1}}{15}$
$I_{2} = \frac{5 \times 2}{15}$
$I_{2} = \frac{2}{3} A$
Hence, heat in $2 Ω$ resistor per second,
$\frac{H}{t} = {(I_{1} + I_{2})}^{2} \times 2$
$\frac{H}{t} = {(2 + \frac{2}{3})}^{2} \times 2$
$\frac{H}{t} = {(\frac{8}{3})}^{2} \times 2$
$\frac{H}{t} = 14.2 cal / s$

AIIMS-2012

Current Electricity

152747 Assertion: Long distance power transmission is done at high voltage.
Reason: At high voltage supply power losses are less.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A As to transmit power over long distance the transmission is done high voltages supply power losses are less.
$P_{loss} = I^{2} R = {(\frac{P}{V})}^{2} R$
Power loss is inversely proportional to the square of the voltage.

AIIMS-2014

Current Electricity

152748 Assertion: When current through a bulb decreases by $0.5 %$ , the glow of bulb decreases by $1 %$ .
Reason: Glow (Power) which is directly proportional to square of current.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A We know that,
Power, $P = I^{2} R$
$\frac{dP}{P} = 2 \frac{dI}{I}$
$= 2 \times 0.5 % = 1 %$
Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 %$ when current through the bulb decreases by $0.5 %$ .

AIIMS-2015

Current Electricity

152750 A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if

1 both the length and radius of wire are halved

2 both length and radius of wire are doubled

3 the radius of wire is doubled

4 the length of the wire is doubled.

Explanation:

B The heat produced is given by,
$H = \frac{V^{2}}{R} t$
We know that,
$R = \frac{ρ l}{π r^{2}}$
Where, Length of wire $= l$
$Radius of wire = r$
$ρ = resistivity$
On putting the value $R$ in equation (i), We get -
$Then, H = \frac{V^{2}}{\frac{ρ l}{π r^{2}}} t$
$H = V^{2} (\frac{π r^{2}}{ρ l}) t$
When, $l$ and $r$ are doubled.
Then, $H^{'} = \frac{V^{2} π}{ρ} (\frac{(2 r)^{2}}{2 l}) t$
$H^{'} = \frac{V^{2}}{ρ} (\frac{π r^{2}}{l}) t \times 2$
$H^{'} = 2 H$
So, heat produced will be doubled when both the length and radius of the wire are doubled.

AIIMS-2012

Current Electricity

152749 Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on.
Reason: Dimness decreases after sometime.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

AIIMS-2008

Current Electricity

152751 In the circuit shown in figure, the $5 Ω$ resistance develops $20.00 cal / s$ due to the current flowing through it. The heat developed in $2 Ω$ resistance (in $cal / s$ ) is

1 23.8

2 14.2

3 11.9

4 7.1

Explanation:

B Given, original image
$Heat, H = I^{2} Rt$
$\frac{H}{t} = I^{2} R$
For $R = 5 Ω$ , and $\frac{H}{t} = 20.00 cal / s$
$∴ 20 = I_{1}^{2} \times 5$
$I_{1} = 2 A$
Voltage across $5 Ω =$ Voltage across $(6 + 9) Ω$
Now, $V = I_{1} \times 5 = I_{2} \times (6 + 9)$
$5 I_{1} = 15 I_{2}$
$∴ I_{2} = \frac{5 I_{1}}{15}$
$I_{2} = \frac{5 \times 2}{15}$
$I_{2} = \frac{2}{3} A$
Hence, heat in $2 Ω$ resistor per second,
$\frac{H}{t} = {(I_{1} + I_{2})}^{2} \times 2$
$\frac{H}{t} = {(2 + \frac{2}{3})}^{2} \times 2$
$\frac{H}{t} = {(\frac{8}{3})}^{2} \times 2$
$\frac{H}{t} = 14.2 cal / s$

AIIMS-2012