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

Answer 1

Answer: (A)

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