Impedance Of Series Lr Circuit
A series circuit contains a resistor of 30 ohm and an inductor whose inductive reactance is 40 ohm, connected to an alternating source. What is the total impedance offered by this combination to the alternating current?
Select the correct option:
Solution
50 ohm
As presented in NCERT Class 12, Chapter 7 (Alternating Current), in a series LR circuit the resistance and inductive reactance are perpendicular in the phasor diagram, so the impedance is their vector sum: Z=R2+XL2, not their simple algebraic sum. Substituting the values: Z=302+402=900+1600=2500=50 Ω. The option 70 ohm incorrectly adds resistance and reactance arithmetically, ignoring their ninety-degree phase difference. The option 10 ohm subtracts them, which would apply only to opposing reactances, not to resistance and reactance. The option 35 ohm simply averages the two values, which has no physical basis. A plausibility check confirms that impedance must be larger than either individual component but smaller than their arithmetic sum, and 50 ohm lies correctly between 40 and 70 ohm, consistent with the Pythagorean combination of perpendicular phasors. The physical reason for the perpendicular combination is that the voltage across a pure resistor is in phase with the current, while the voltage across a pure inductor leads the current by ninety degrees, so their voltage phasors add at right angles. This same impedance also determines the phase angle of the circuit through tanϕ=XL/R=40/30, giving a phase angle whose cosine is the power factor R/Z=30/50=0.6. Recognising impedance as a phasor sum rather than a scalar sum is one of the most common conceptual hurdles in alternating-current problems, and mastering it is essential for analysing any reactive circuit.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- impedance of series lr circuit
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
50 ohm
As presented in NCERT Class 12, Chapter 7 (Alternating Current), in a series LR circuit the resistance and inductive reactance are perpendicular in the phasor diagram, so the impedance is their vector sum: Z=R2+XL2, not their simple algebraic sum. Substituting the values: Z=302+402=900+1600=2500=50 Ω. The option 70 ohm incorrectly adds resistance and reactance arithmetically, ignoring their ninety-degree phase difference. The option 10 ohm subtracts them, which would apply only to opposing reactances, not to resistance and reactance. The option 35 ohm simply averages the two values, which has no physical basis. A plausibility check confirms that impedance must be larger than either individual component but smaller than their arithmetic sum, and 50 ohm lies correctly between 40 and 70 ohm, consistent with the Pythagorean combination of perpendicular phasors. The physical reason for the perpendicular combination is that the voltage across a pure resistor is in phase with the current, while the voltage across a pure inductor leads the current by ninety degrees, so their voltage phasors add at right angles. This same impedance also determines the phase angle of the circuit through tanϕ=XL/R=40/30, giving a phase angle whose cosine is the power factor R/Z=30/50=0.6. Recognising impedance as a phasor sum rather than a scalar sum is one of the most common conceptual hurdles in alternating-current problems, and mastering it is essential for analysing any reactive circuit.
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of impedance of series lr circuit. It appeared in the 2025 exam.
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