A resistor of 20 Omega and a capacitor are c | vedclass

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Question

(B) The phase angle $\phi$ in an $RC$ series circuit is given by $	an(\phi) = \frac{X_C}{R}$.Given $\phi = 30^\circ$,$R = 20 \ \Omega$,and $f = 50 \

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$\frac{\sqrt{3}}{2 \pi} \ mF$

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(B) The phase angle $\phi$ in an $RC$ series circuit is given by $	an(\phi) = \frac{X_C}{R}$.Given $\phi = 30^\circ$,$R = 20 \ \Omega$,and $f = 50 \ Hz$.$	an(30^\circ) = \frac{1}{\sqrt{3}} = \frac{X_C}{20}$.Therefore,$X_C = \frac{20}{\sqrt{3}} \ \Omega$.Since $X_C = \frac{1}{2 \pi f C}$,we have $\frac{1}{2 \pi 	imes 50 	imes C} = \frac{20}{\sqrt{3}}$.$\frac{1}{100 \pi C} = \frac{20}{\sqrt{3}}$.Solving for $C$: $C = \frac{\sqrt{3}}{2000 \pi} \ F$.To convert to $mF$ (millifarads),multiply by $10^3$: $C = \frac{\sqrt{3}}{2000 \pi} 	imes 10^3 \ mF = \frac{\sqrt{3}}{2 \pi} \ mF$.

$A$ resistor of $20 \ \Omega$ and a capacitor are connected in series with an $AC$ current source of $50 \ Hz$. What should be the capacitance to produce a phase difference of $30^\circ$ between the voltage and current?

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