In the phasor diagram of a series LCR circuit, the phasor V of the source voltage is the hypotenuse of a right triangle. Here \(v_m\), \(v_{Rm}\), \(v_{Lm}\) and \(v_{Cm}\) are the amplitudes of the voltages across the source, R, L and C. Which of the following relations follows from this triangle?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(v_m^{2}=v_{Rm}^{2}+(v_{Cm}-v_{Lm})^{2}\)
In a series circuit the same current flows through R, L and C. The phasor \( V_R \) is along the current phasor. \( V_L \) is \( \pi/2 \) ahead of it and \( V_C \) is \( \pi/2 \) behind it.
Option (1) \(v_m^{2}=v_{Rm}^{2}+(v_{Cm}+v_{Lm})^{2}\) — Incorrect. The phasors \( V_C \) and \( V_L \) point in opposite directions. So the length of their combined phasor is the difference \( |v_{Cm}-v_{Lm}| \).
Option (2) \(v_m^{2}=v_{Rm}^{2}+(v_{Cm}-v_{Lm})^{2}\) — Correct. \( V_C \) and \( V_L \) lie on one line and combine into one phasor of length \( |v_{Cm}-v_{Lm}| \). This phasor is at right angles to \( V_R \). With V as the hypotenuse, the Pythagorean theorem gives this relation.
Option (3) \(v_m^{2}=v_{Rm}^{2}+v_{Lm}^{2}+v_{Cm}^{2}\) — Incorrect. This would need all three phasors at right angles to one another. But \( V_L \) and \( V_C \) are along the same line.
Option (4) \(v_m=v_{Rm}+|v_{Cm}-v_{Lm}|\) — Incorrect. Voltages that differ in phase cannot be added like ordinary numbers. Two sides at right angles combine by the Pythagorean theorem.
NCERT (2026-27):
Phasor-diagram solution – “… Since VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude |vCm – vLm|. Since V is represented as the hypotenuse of a right-triangle whose sides are VR and (VC + VL), the pythagorean theorem gives:”
\( v_m^{2}=v_{Rm}^{2}+(v_{Cm}-v_{Lm})^{2} \)