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Question: If the current I through a resistor is increased by \( 100\% \) (assume that temperature remains unc...

If the current I through a resistor is increased by 100%100\% (assume that temperature remains unchanged), the increase in power dissipated will be
A) 100%100\%
B) 200%200\%
C) 300%300\%
D) 400%400\%

Explanation

Solution

The power dissipated through a resistor is proportional to the square of the current flowing through it. Find the power dissipated in the circuit when the current is increased by 100%100\% and compare it with the power dissipated to find the increase from the former case.

Formula used: P=I2RP = {I^2}R where PP is the power dissipated through a resistor, II is the current flowing through it and RR is the resistance.

Complete step by step solution:
When a current II is flowing through a resistor of resistance RR , the power dissipated across it can be calculated as P=I2RP = {I^2}R .
Now, when the current through the resistor is increased by 100%100\% , the new value of current in the circuit denoted by II' can be calculated as
I=I+100%ofI\Rightarrow I' = I + 100\% \,{\text{of}}\,I
I=2I\Rightarrow I' = 2I
Then the power dissipated across the resistor can be calculated as:
P=I2R\Rightarrow P' = I{'^2}R
Substituting I=2II' = 2I in the above equation, we can rewrite PP' as:
P=4I2R\Rightarrow P' = 4{I^2}R
P=4P\Rightarrow P' = 4P
The change in power when the current is increased by 100%100\% can be determined as:
Increase in power dissipated is PPP×100\dfrac{{P' - P}}{P} \times 100
Substituting P=4PP' = 4P in the increase in power dissipated, we can write:
Increase in power dissipated is 4PPP×100=300%\dfrac{{4P - P}}{P} \times 100 = 300\%
Hence, on increasing the current by 100%100\% , the power dissipated across the resistor increases by 300%300\% which corresponds to (C).
Hence, the correct option is option (C).

Note:
The power dissipated across the resistor here has been calculated assuming the temperature and hence and resistance of the resistor remain constant. In reality, however, higher power dissipated across the resistor can heat it and increase in resistance which would then affect the current flowing in the circuit.