Question

In a series RLC circuit, L = 40 mH is given. If the instantaneous voltage and current are $100\cos (314t-5^{\circ })V$ and $10\cos (314t-{50}^{\circ })A$ respectively, the value of R and C will be

• A. $R=10\Omega andC=580\mu F$
• B. $R=7.07\Omega andC=580\mu F$
• C. $R=7.07\Omega andC=5.49mF$
• D. $R=14.14\Omega andC=5.49mF$

Answer 1

The correct option is B $R=7.07\Omega andC=580\mu F$

The current lags the voltage by ${50}^{\circ }–5^{\circ }={45}^{\circ }$

$\therefore \omega L>\frac{1}{\omega C}$

$\tan {45}^{\circ }=1=\frac{\omega L-\frac{1}{\omega C}}{R}$

and $\frac{V_m}{I_m}=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}=\sqrt{R^2+R^2}$

$\frac{100}{10}=\sqrt{2}R$

or $R=7.07\Omega$

$\because R=\omega L-\frac{1}{\omega C}$

$\frac{1}{\omega C}=314\times 40\times {10}^{-3}-7.07$

$\frac{1}{\omega C}=12.56-7.07=5.49$

or $C=\frac{1}{314\times 5.49}\approx 580\mu F$