Question

The series combination of a Capacitor C and an Inductor (L = 1.0 H) is connected across an alternating voltage source of 220 V (rms) at a frequency of 50Hz. An ideal voltmeter is connected across the capacitor. Determine the value of the capacitance C such that the voltmeter reads 220 Volts. [Use $\pi^2$=10]

Answer 1

20 μF

Answer 2

The circuit consists of a capacitor (C) and an inductor (L) connected in series across an alternating voltage source. An ideal voltmeter is connected across the capacitor.

Given:

• Inductance, $L = 1.0$ H
• RMS voltage of the source, $V_{rms} = 220$ V
• Frequency of the source, $f = 50$ Hz
• Voltmeter reading across the capacitor, $V_C = 220$ V
• Use $\pi^2 = 10$

In a series LC circuit, the RMS voltage across the source ($V_{rms}$) is the magnitude of the difference between the RMS voltage across the inductor ($V_L$) and the RMS voltage across the capacitor ($V_C$), because these voltages are 180 degrees out of phase. The formula is: $V_{rms} = |V_L - V_C|$

Substitute the given values into the formula: $220 = |V_L - 220|$

This equation yields two possibilities for $V_L$:

1. $V_L - 220 = 220 \implies V_L = 440$ V
2. $V_L - 220 = -220 \implies V_L = 0$ V

Since the inductor has a non-zero inductance ($L = 1.0$ H) and the frequency is non-zero ($f = 50$ Hz), the inductive reactance $X_L = 2\pi f L$ will be non-zero. If $V_L$ were 0, it would imply either $X_L=0$ (which is not true) or the current $I_{rms}=0$. If $I_{rms}=0$, then $V_C = I_{rms} X_C$ would also be 0, which contradicts the given $V_C = 220$ V. Therefore, $V_L = 0$ V is not physically possible in this scenario. So, we must have $V_L = 440$ V.

In a series circuit, the RMS current ($I_{rms}$) is the same through both components. We can express the voltages across the components in terms of current and reactance: $V_C = I_{rms} X_C$ $V_L = I_{rms} X_L$

From these, we can write: $I_{rms} = \frac{V_C}{X_C} = \frac{V_L}{X_L}$

Substitute the values of $V_C$ and $V_L$: $\frac{220}{X_C} = \frac{440}{X_L}$

This simplifies to: $X_L = 2 X_C$

Now, we substitute the formulas for inductive reactance ($X_L$) and capacitive reactance ($X_C$): $X_L = \omega L = 2 \pi f L$ $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$

Substitute these into the relationship $X_L = 2 X_C$: $2 \pi f L = 2 \left( \frac{1}{2 \pi f C} \right)$ $2 \pi f L = \frac{1}{\pi f C}$

Now, solve for the capacitance C: $C = \frac{1}{2 \pi f L \cdot \pi f}$ $C = \frac{1}{2 \pi^2 f^2 L}$

Substitute the given numerical values: $f = 50$ Hz $L = 1.0$ H $\pi^2 = 10$

$C = \frac{1}{2 \times 10 \times (50)^2 \times 1.0}$ $C = \frac{1}{20 \times 2500}$ $C = \frac{1}{50000}$ $C = \frac{1}{5 \times 10^4}$ $C = 0.2 \times 10^{-4}$ F $C = 2 \times 10^{-5}$ F $C = 20 \times 10^{-6}$ F $C = 20 \, \mu F$

The value of the capacitance C is $20 \, \mu F$.