Question:
An alternating supply of \(225\,\text{V}\) is applied across a circuit with resistance \(20\,\Omega\) and impedance \(45\,\Omega\). The power dissipated in the circuit is
An alternating supply of \(225\,\text{V}\) is applied across a circuit with resistance \(20\,\Omega\) and impedance \(45\,\Omega\). The power dissipated in the circuit is
Show Hint
For an AC circuit,
\[
I=\frac{V}{Z}
\]
and the actual power consumed is
\[
P=I^2R
=
VI\cos\phi.
\]
Only the resistive part of the circuit dissipates energy.
Updated On: Jul 9, 2026
- \(500\,\text{W}\)
- \(1000\,\text{W}\)
- \(550\,\text{W}\)
- \(2100\,\text{W}\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
In an AC circuit,
\[
P=VI\cos\phi.
\]
Since
\[
\cos\phi=\frac{R}{Z},
\]
the average power can also be written as
\[
P=\frac{V^2R}{Z^2},
\]
where
\[
V=\text{rms voltage},
\quad
R=\text{resistance},
\quad
Z=\text{impedance}.
\]
Step 1: Calculate the current in the circuit. \[ I=\frac{V}{Z} =\frac{225}{45} =5\,\text{A}. \]
Step 2: Calculate the power dissipated. Only the resistor dissipates power. \[ P=I^2R. \] Substituting, \[ P=(5)^2(20). \] \[ P=25\times20. \] \[ P=500\,\text{W}. \]
Step 3: Write the final answer. \[ \boxed{P=500\,\text{W}} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Calculate the current in the circuit. \[ I=\frac{V}{Z} =\frac{225}{45} =5\,\text{A}. \]
Step 2: Calculate the power dissipated. Only the resistor dissipates power. \[ P=I^2R. \] Substituting, \[ P=(5)^2(20). \] \[ P=25\times20. \] \[ P=500\,\text{W}. \]
Step 3: Write the final answer. \[ \boxed{P=500\,\text{W}} \] \[ \boxed{\text{Answer = (A)}} \]
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