Question

An inductor of inductance $L,$ a capacitor of capacitance $C$ and a resistor of resistance $R$ are connected in series to an ac source of potential difference $V\text{volt}$ as shown in figure. Potential difference across $L,C$ and $R$ is $40\text{V},10\text{V}$ and $40\text{V},$ respectively. The amplitude of current flowing through $LCR$ series circuit is $10\sqrt{2}\text{A}.$ The impedance of the circuit is

• A. $5\Omega$
• B. $4\sqrt{2}\Omega$
• C. $5\sqrt{2}\Omega$
• D. $4\Omega$

Answer 1

The correct option is A $5\Omega$

Given:
$V_L=40\text{volt};V_R=40\text{volt};V_C=10\text{volt}$

Now,
$V_{\text{RMS}}=\sqrt{V_R^2+(V_L-V_C{)}^2}$

$=\sqrt{(40{)}^2+(40-10{)}^2}=50\text{V}$

$I_{\text{RMS}}=\frac{I_0}{\sqrt{2}}=\frac{10\sqrt{2}}{\sqrt{2}}=10\text{A}$

$\because V_{\text{RMS}}=I_{\text{RMS}}\times Z$

$\therefore Z=\frac{V_{\text{RMS}}}{I_{\text{RMS}}}=\frac{50}{10}=5\Omega$

Hence, option $\left(\text{A}\right)$ is correct.