Question

A bulb is connected in series with a capacitor to an AC supply. If the capacitance of the capacitor increases, then power of light emitted by the bulb

• A. decreases
• B. increases
• C. does not change
• D. becomes zero

Hint:

In an AC circuit with a capacitor, remember that capacitance is inversely proportional to capacitive reactance ($X_C \propto 1/C$). A decrease in $X_C$ leads to a decrease in overall circuit impedance for a series R-C circuit, resulting in an increased current and, consequently, increased power dissipation in the resistive component (the bulb).

Answer 1

The Correct Option is B

Step 1: Understand the Circuit and Components
The circuit consists of a bulb connected in series with a capacitor to an AC supply. \begin{itemize} \item Bulb: A bulb primarily acts as a resistor in an AC circuit. The brightness (power of light emitted) of the bulb depends on the current flowing through it. Higher current means more power dissipated by the bulb, hence brighter light. \item Capacitor: In an AC circuit, a capacitor offers capacitive reactance ($X_C$), which is an opposition to the flow of alternating current. \end{itemize} Step 2: Recall the Formula for Capacitive Reactance
The capacitive reactance ($X_C$) of a capacitor in an AC circuit is given by: \[ X_C = \frac{1}{2\pi f C} \] Where: \begin{itemize} \item $X_C$ is the capacitive reactance (measured in ohms). \item $f$ is the frequency of the AC supply. \item $C$ is the capacitance of the capacitor. \end{itemize} Step 3: Analyze the Effect of Increasing Capacitance on Reactance
From the formula for $X_C$, we can see that capacitive reactance is inversely proportional to capacitance ($C$). \[ X_C \propto \frac{1}{C} \] Therefore, if the capacitance ($C$) of the capacitor increases, the capacitive reactance ($X_C$) will decrease. Step 4: Analyze the Effect of Decreasing Reactance on Circuit Impedance and Current
The total opposition to current flow in an AC series circuit containing a resistor (bulb) and a capacitor is called impedance ($Z$). The impedance is given by: \[ Z = \sqrt{R^2 + X_C^2} \] Where $R$ is the resistance of the bulb. Since the capacitive reactance ($X_C$) decreases (as established in Step 3) and the resistance ($R$) of the bulb remains constant, the total impedance ($Z$) of the circuit will decrease. Now, consider the current ($I$) flowing through the circuit. According to Ohm's Law for AC circuits: \[ I = \frac{V}{Z} \] Where $V$ is the voltage of the AC supply. Since the supply voltage ($V$) is constant and the total impedance ($Z$) of the circuit decreases, the current ($I$) flowing through the circuit will increase. Step 5: Relate Current to Power of Light Emitted by the Bulb
The power ($P$) dissipated by the bulb (and thus the power of light emitted) is given by: \[ P = I^2 R \] Since the current ($I$) flowing through the bulb increases (as established in Step 4) and the resistance ($R$) of the bulb is constant, the power ($P$) of light emitted by the bulb will increase. Step 6: Analyze Options
\begin{itemize} \item Option (1): decreases. Incorrect, as the power increases. \item Option (2): increases. Correct, as the power increases. \item Option (3): does not change. Incorrect. \item Option (4): becomes zero. Incorrect. \end{itemize}