A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:
- increases
- remains same
- becomes zero
- decreases
The Correct Option is A
Approach Solution - 1
To solve this problem, we need to understand how a capacitor affects the circuit when connected in series with a bulb and how inserting a dielectric affects the capacitor.
Theoretical Background:
- A capacitor introduces capacitive reactance in an AC circuit.
- The capacitive reactance \(X_C\) is given by the formula: \(X_C = \frac{1}{2\pi fC}\), where \(f\) is the frequency of the AC supply, and \(C\) is the capacitance.
- Inserting a dielectric between the plates of a capacitor increases its capacitance \((C = \kappa C_0)\) as \(\kappa\) is the dielectric constant, which is greater than 1.
Effect of Dielectric on the Circuit:
- With an increased capacitance \((C)\), the capacitive reactance \((X_C)\) decreases.
- Lower reactance means less impedance in the circuit, allowing more current to flow through the bulb.
Conclusion:
As a result, the glow of the bulb increases because more current flows through it due to the lowered impedance caused by the increased capacitance from inserting the dielectric.
Justification of Options:
- Increases: Correct, due to reduced reactance and increased current flow.
- Remains same: Incorrect, as changing capacitance affects current flow.
- Becomes zero: Incorrect, as increased capacitance increases current, not stops it.
- Decreases: Incorrect, increased capacitance decreases impedance, thus increasing current.
Approach Solution -2
Step 1: Understanding the circuit The capacitor is in series with the bulb in an AC circuit. The impedance Z of the capacitor is given by:
\[ Z_C = \frac{1}{\omega C}, \]
where:
- \(\omega\) is the angular frequency of the AC supply,
- \(C\) is the capacitance of the capacitor.
Step 2: Effect of placing a dielectric Placing a dielectric between the plates of the capacitor increases the capacitance \(C\), as:
\[ C' = \kappa C, \]
where \(\kappa > 1\) is the dielectric constant.
Step 3: Impedance of the capacitor Since \(Z_C \propto \frac{1}{C}\), increasing \(C\) reduces the capacitive impedance \(Z_C\).
Step 4: Impact on the bulb The total impedance of the circuit decreases, leading to an increase in the current through the circuit. As the current increases, the glow of the bulb increases.
Final Answer: The glow of the bulb increases.
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