Question

A capacitor and a resistor of resistance $100\sqrt{3}\Omega$ are connected in series to an ac source of voltage $100\sin(200t)$ V, where 't' is time in second. If the phase difference between the voltage and the current in the circuit is 30$^\circ$, then the capacitance of the capacitor is

• A. 30 $\mu$F
• B. 50 $\mu$F
• C. 100 $\mu$F
• D. 150 $\mu$F

Hint:

For a series RC circuit, remember the impedance triangle, which gives the relationship $\tan\phi = X_C/R$. Also, identify the angular frequency $\omega$ from the AC voltage or current equation of the form $V_0\sin(\omega t)$.

Answer 1

The Correct Option is B

In a series RC circuit, the phase difference ($\phi$) between the voltage and current is given by:
$\tan\phi = \frac{X_C}{R}$.
Where $X_C$ is the capacitive reactance and $R$ is the resistance.
We are given:
Resistance, $R = 100\sqrt{3} \Omega$.
Phase difference, $\phi = 30^\circ$.
We can find the capacitive reactance $X_C$:
$\tan(30^\circ) = \frac{X_C}{100\sqrt{3}}$.
$\frac{1}{\sqrt{3}} = \frac{X_C}{100\sqrt{3}}$.
$X_C = \frac{100\sqrt{3}}{\sqrt{3}} = 100 \Omega$.
The capacitive reactance is also given by the formula $X_C = \frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance.
The voltage source is given by $V = 100\sin(200t)$. Comparing this with the standard form $V=V_0\sin(\omega t)$, we can identify the angular frequency:
$\omega = 200$ rad/s.
Now we can solve for the capacitance $C$:
$100 = \frac{1}{200 \cdot C}$.
$C = \frac{1}{100 \times 200} = \frac{1}{20000} = \frac{1}{2 \times 10^4}$ F.
$C = 0.5 \times 10^{-4}$ F = $5 \times 10^{-5}$ F.
To express this in microfarads ($\mu$F), we use $1 \mu\text{F} = 10^{-6}$ F.
$C = 50 \times 10^{-6}$ F = 50 $\mu$F.