AIIMS 2005 Physics Voltage in AC Circuits MCQ Question

To solve the problem, we need to determine the voltage across the capacitor $C$ in an AC circuit where a 20 V (rms) AC source is connected across a resistor $R$ and a capacitor $C$. The voltage across the resistor $R$ is given as 12 V.

Step 1: Understanding the Circuit

In an AC circuit with a resistor $R$ and capacitor $C$ in series, the total voltage $V$ is the vector sum of the voltages across the resistor $V_R$ and the capacitor $V_C$.

Using the concept of phasors, we know that:

• The voltage across the resistor $V_R$ is in phase with the current.
• The voltage across the capacitor $V_C$ lags the current by $90^\circ$.

Step 2: Applying the Pythagorean Theorem

Since the voltages are out of phase, we can use the Pythagorean theorem to relate the total voltage $V$, the voltage across the resistor $V_R$, and the voltage across the capacitor $V_C$:

$$V^2 = V_R^2 + V_C^2$$

Step 3: Plugging in Values

We know:

• The total voltage $V = 20 \, \text{V}$
• The voltage across the resistor $V_R = 12 \, \text{V}$

Substituting these values into the equation gives:

$$20^2 = 12^2 + V_C^2$$

Calculating $20^2$ and $12^2$:

$$400 = 144 + V_C^2$$

Step 4: Solving for $V_C$

Now, isolate $V_C^2$:

$$V_C^2 = 400 - 144 = 256$$

Taking the square root to find $V_C$:

$$V_C = \sqrt{256} = 16 \, \text{V}$$

Thus, the voltage across the capacitor $C$ is $16 \, \text{V}$.

Conclusion

The correct answer is B) 16 V.

Clarifying Other Options

• A) 8 V: This value does not satisfy the Pythagorean relationship established between the voltages.
• C) 10 V: This value also fails to fulfill the condition since substituting $V_C = 10 \, \text{V}$ does not satisfy the equation $400 = 144 + 100$.
• D) Not possible to determine unless values of R and C are given: This option is incorrect because we can determine the voltage across the capacitor with the given voltage across the resistor and the total voltage.

Thus, the only valid conclusion based on the provided information is that the voltage across the capacitor $C$ is indeed $16 \, \text{V}$.