Question

For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414\, \Omega$, then the values of $P$ and $Q$ are

• A. $1\, k \Omega ,\, 10\, \mu F $
• B. $1\, k \Omega ,\, 1\, \mu F $
• C. $1\, k \Omega , 10\, \, mH$
• D. $1\, k \Omega ,\, 1\, mH$

Answer 1

Answer: (A)

In the shown figure, current is ahead of voltage, so its a $R C$ circuit, so $P$ is a resistor and $Q$ is a capacitor.
Now, in $R C$ circuit,
$Z=\sqrt{R^{2}+X_{C}^{2}}$
For simple solution,
Given $Z=1414=1000 \times 1.414$
$=1000 \sqrt{2}$
$=1000 \sqrt{1+1}$
$=\sqrt{(1000)^{2}+(1000)^{2}}$
$\Rightarrow X_{C}=1000 R=1000\, \Omega$
Now, $X_{C}=\frac{1}{\omega C}$
$\Rightarrow C=\frac{1}{\omega X_{C}}$
$\Rightarrow C=\frac{1}{100 \times 1000}=10\, \mu F$