If the current I through a resistor is increased by $ 100\% $ (assume that temperature remains unchanged), the increase in power dissipated will be
A) $ 100\% $
B) $ 200\% $
C) $ 300\% $
D) $ 400\% $
Answer
550.3k+ views
Hint: 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\% $ and compare it with the power dissipated to find the increase from the former case.
Formula used: $ P = {I^2}R $ where $ P $ is the power dissipated through a resistor, $ I $ is the current flowing through it and $ R $ is the resistance.
Complete step by step solution:
When a current $ I $ is flowing through a resistor of resistance $ R $ , the power dissipated across it can be calculated as $ P = {I^2}R $ .
Now, when the current through the resistor is increased by $ 100\% $ , the new value of current in the circuit denoted by $ I' $ can be calculated as
$\Rightarrow I' = I + 100\% \,{\text{of}}\,I $
$\Rightarrow I' = 2I $
Then the power dissipated across the resistor can be calculated as:
$\Rightarrow P' = I{'^2}R $
Substituting $ I' = 2I $ in the above equation, we can rewrite $ P' $ as:
$\Rightarrow P' = 4{I^2}R$
$\Rightarrow P' = 4P $
The change in power when the current is increased by $ 100\% $ can be determined as:
Increase in power dissipated is $ \dfrac{{P' - P}}{P} \times 100 $
Substituting $ P' = 4P $ in the increase in power dissipated, we can write:
Increase in power dissipated is $ \dfrac{{4P - P}}{P} \times 100 = 300\% $
Hence, on increasing the current by $ 100\% $ , the power dissipated across the resistor increases by $ 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.
Formula used: $ P = {I^2}R $ where $ P $ is the power dissipated through a resistor, $ I $ is the current flowing through it and $ R $ is the resistance.
Complete step by step solution:
When a current $ I $ is flowing through a resistor of resistance $ R $ , the power dissipated across it can be calculated as $ P = {I^2}R $ .
Now, when the current through the resistor is increased by $ 100\% $ , the new value of current in the circuit denoted by $ I' $ can be calculated as
$\Rightarrow I' = I + 100\% \,{\text{of}}\,I $
$\Rightarrow I' = 2I $
Then the power dissipated across the resistor can be calculated as:
$\Rightarrow P' = I{'^2}R $
Substituting $ I' = 2I $ in the above equation, we can rewrite $ P' $ as:
$\Rightarrow P' = 4{I^2}R$
$\Rightarrow P' = 4P $
The change in power when the current is increased by $ 100\% $ can be determined as:
Increase in power dissipated is $ \dfrac{{P' - P}}{P} \times 100 $
Substituting $ P' = 4P $ in the increase in power dissipated, we can write:
Increase in power dissipated is $ \dfrac{{4P - P}}{P} \times 100 = 300\% $
Hence, on increasing the current by $ 100\% $ , the power dissipated across the resistor increases by $ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How many of the following diseases can be studied with class 11 biology CBSE

Which of the following enzymes is used for carboxylation class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

