Question

To a highly inductive circuit, a small capacitance is added in series. The angle between voltage and current will,

• A. increase
• B. decrease
• C. remain nearly same
• D. become indeterminant

Hint:

In series AC circuits, inductive and capacitive reactances oppose each other. Adding capacitance to an inductive circuit always reduces the phase angle.

Answer 1

The Correct Option is B

Step 1: Understand the nature of a highly inductive circuit.
In a highly inductive circuit, the current lags the voltage by a large angle because the inductive reactance dominates over resistance.
Step 2: Effect of adding a capacitor in series.
When a capacitor is added in series, it introduces capacitive reactance which opposes the inductive reactance.
This reduces the net reactive component of the circuit impedance.
Step 3: Effect on phase angle.
The phase angle between voltage and current depends on the ratio of net reactance to resistance.
Since the net reactance decreases due to partial cancellation of inductive reactance by capacitive reactance, the phase angle also decreases.
Step 4: Conclusion.
Therefore, adding a small capacitance in series with a highly inductive circuit causes the angle between voltage and current to
\[ \boxed{\text{decrease}} \]

Answer 2

In a highly inductive series circuit, the impedance is dominated by inductive reactance, so the phase angle \(\phi = \tan^{-1}(X_L/R)\) is large and the current lags the voltage substantially. Adding a small capacitance in series introduces a capacitive reactance \(X_C\) that subtracts from \(X_L\), since inductive and capacitive reactances have opposite sign in a series circuit. Checking each option against this net-reactance picture:

1. Increase: For the angle to increase, the net reactance \(X_L - X_C\) would have to grow larger than \(X_L\) alone, which cannot happen when a capacitor (which subtracts reactance) is added.
2. Decrease: Since \(X_C\) subtracts from \(X_L\), the net series reactance \(X_L - X_C\) is smaller than \(X_L\) alone, so \(\tan^{-1}\!\big((X_L-X_C)/R\big)\) is smaller than the original angle. The phase angle decreases.
3. Remain nearly same: This would only hold if \(X_C\) were negligibly small compared to \(X_L\), but the question specifies the added capacitance is enough to matter, so a genuine (if partial) reduction occurs.
4. Become indeterminate: The direction of the effect is fixed and predictable from the series reactance subtraction; it does not become indeterminate just because a component is added.

Adding a series capacitance to a highly inductive circuit always works against the inductive reactance, pulling the net reactance down and the phase angle down with it.

Therefore, the correct answer is decrease.