In a series R-L-C circuit at resonance, the voltages across the resistor and the inductor are 30V and 40V…

Analysis

This question tests your understanding of the properties of a series R-L-C circuit at resonance. You are given the voltage across the resistor ($V_R$) and the inductor ($V_L$), and you need to find the voltage across the capacitor ($V_C$).

Concept

In a series R-L-C circuit, resonance occurs when the inductive reactance ($X_L$) is equal to the capacitive reactance ($X_C$).

Mathematically: $X_L = X_C$

Since the current ($I$) flowing through all components in a series circuit is the same, we can multiply both sides of the equation by $I$: $I \times X_L = I \times X_C$

This leads to the fundamental condition for resonance in terms of voltage: $V_L = V_C$

This means that at resonance, the voltage across the inductor is exactly equal in magnitude to the voltage across the capacitor.

Solution

1. Identify the given values: • Voltage across the resistor ($V_R$) = 30V • Voltage across the inductor ($V_L$) = 40V

2. Apply the resonance condition: • At resonance, the inductive voltage and capacitive voltage are equal ($V_L = V_C$). • Since $V_L = 40V$, then $V_C$ must also be 40V.

Note: The voltage across the resistor ($V_R$) is equal to the total supply voltage of the circuit at resonance, but it does not change the fact that $V_L$ must equal $V_C$.

Conclusion

Based on the principle of resonance in R-L-C circuits, the voltage across the capacitor is 40V.

Exam Tip

Always remember that at resonance, the circuit behaves as if it were purely resistive. The voltages across the inductor and capacitor are $180^\circ$ out of phase, which is why they "cancel" each other out in the total voltage calculation ($V_{total} = \sqrt{V_R^2 + (V_L - V_C)^2}$). Since $V_L - V_C = 0$, the total voltage is simply $V_R$.