Question

An $\text{RLC}$ series circuit has $L=10\text{mH},R=3\Omega$ and $C=1\mu \text{F}$ connected in series to a source of $E=15\cos \omega t\text{V}$. If the frequency of the voltage is $10\%$ less than the resonant frequency, then

• A. Peak value of current is $0.704\text{A}$
• B. Impedance of the circuit is $21.32\Omega$
• C. Power factor of the circuit is $0.141$
• D. Resonant frequency of the circuit is $9\times {10}^3\text{rad/s}$

Answer 1

Answer: (A), (B), (C)

Power factor of the circuit is $0.141$

Given:-

$L=10\text{mH};C=1\mu \text{F}$

$R=3\Omega ;E=15cos\omega t$

Resonant frequency of the circuit is given as

${\omega }_R=\frac{1}{\sqrt{LC}}$

${\omega }_R=\frac{1}{\sqrt{10\times {10}^{-3}\times {10}^{-6}}}$

${\omega }_R={10}^4\text{rad/s}$

Now, $10\%$ less frequency than the resonant frequency is given as

$\omega ={10}^4-\frac{10}{100}\times {10}^4=9\times {10}^3\text{rad/s}$

At this frequency

$X_L=\omega L=9\times {10}^3\times (10\times {10}^{-3})$

$X_L=90\Omega$

and $X_C=\frac{1}{\omega C}=\frac{1}{9\times {10}^3\times {10}^{-6}}$

$X_C=111.1\Omega$

Impedance of the circuit,

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$Z=\sqrt{R^2+(90-111.1{)}^2}$

$Z=\sqrt{3^2+(21.1{)}^2}$

$\Rightarrow Z=21.32\Omega$

Peak value of current is given as

$I_0=\frac{E_0}{Z}=\frac{15}{21.32}$

$I_0=0.704\text{A}$

Power factor of the circuit is given as

$\cos \varphi =\frac{R}{Z}=\frac{3}{21.32}$

$\Rightarrow \cos \varphi =0.141$

Hence, options $\left(A\right)$,$\left(B\right)$ and $\left(C\right)$ are the correct answers.