Series Lcr Impedance
In a series circuit a resistor of 30 (\Omega) is joined with an inductive reactance of 80 (\Omega) and a capacitive reactance of 40 (\Omega) across an AC source. What is the total impedance presented to the source?
Select the correct option:
Solution
50 \(\Omega\)
In a series LCR circuit the resistor's voltage is in phase with the current while the inductor and capacitor voltages lead and lag by ninety degrees respectively, so they partially cancel; impedance therefore combines resistance and net reactance in quadrature as (Z = \sqrt{R^2 + (X_L - X_C)^2}). The net reactance is (X_L - X_C = 80 - 40 = 40;\Omega), and with (R = 30;\Omega) we obtain (Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50;\Omega). The option 70 (\Omega) wrongly adds reactance to resistance arithmetically. The option 150 (\Omega) sums all three magnitudes linearly, ignoring phase. The option 30 (\Omega) keeps only the resistance, as if at resonance. This is the standard NCERT phasor treatment where reactances subtract before being combined with resistance, since the inductor and capacitor voltage phasors point in opposite directions along the same axis and only their resultant adds in quadrature with the resistive phasor. The phase angle of the circuit is given by (\tan\phi = (X_L - X_C)/R), which here is positive and confirms a current that lags the applied voltage. A plausibility check confirms the impedance exceeds the resistance alone yet remains below the simple arithmetic sum of all three magnitudes, exactly as the 3-4-5 right triangle dictates, and the circuit is net inductive since (X_L > X_C).
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- series lcr impedance
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
50 \(\Omega\)
In a series LCR circuit the resistor's voltage is in phase with the current while the inductor and capacitor voltages lead and lag by ninety degrees respectively, so they partially cancel; impedance therefore combines resistance and net reactance in quadrature as (Z = \sqrt{R^2 + (X_L - X_C)^2}). The net reactance is (X_L - X_C = 80 - 40 = 40;\Omega), and with (R = 30;\Omega) we obtain (Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50;\Omega). The option 70 (\Omega) wrongly adds reactance to resistance arithmetically. The option 150 (\Omega) sums all three magnitudes linearly, ignoring phase. The option 30 (\Omega) keeps only the resistance, as if at resonance. This is the standard NCERT phasor treatment where reactances subtract before being combined with resistance, since the inductor and capacitor voltage phasors point in opposite directions along the same axis and only their resultant adds in quadrature with the resistive phasor. The phase angle of the circuit is given by (\tan\phi = (X_L - X_C)/R), which here is positive and confirms a current that lags the applied voltage. A plausibility check confirms the impedance exceeds the resistance alone yet remains below the simple arithmetic sum of all three magnitudes, exactly as the 3-4-5 right triangle dictates, and the circuit is net inductive since (X_L > X_C).
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of series lcr impedance. It appeared in the 2025 exam.
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