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The potential differences between the two ends of the three components of an L–C–R series A.C. circuit are V_L, V_C and V_R respectively. Then the voltage of the A.C. source is ........

Asked in RS Academy GUJCET booklet · Voltages across the elements

• (1) V_L+V_C+V_R
• (2) V_R+V_L-V_C
• (3) √V_R²+(V_L+V_C)²
• (4) √V_R²+(V_L-V_C)²

Step-by-step solution

Why the other options are wrong

• (1) Adds the three readings as if they were in phase. That works only for instantaneous values taken at the same instant; V_L, V_C and V_R are r.m.s. values of voltages that are out of phase with one another.
• (2) Gets V_L-V_C right but then adds V_R directly. V_R is 90° away from the reactive part, so the two combine as the sides of a right-angled triangle.
• (3) Adds V_L and V_C. They are exactly opposite in phase, 180° apart, so they subtract.

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