The a.c. source is connected to series LCR circuit. If voltage across R is 40 V, that across L is 80 V and that across C is 40 V, then the e.m.f. ' e ' of a.c. source is (a) 40 V (b) 40 √(2) V (c) 80 V (d) 160 V

Answer: (b). e = √(VR² + (VL - VC)²) e = √(40² + (80 - 40)²) = √(3200) = 40√(2) V So the answer is (b) 40√(2) V.

The a.c. source is connected to series LCR circuit. If voltage across… | MHT CET 2023 (11 May, Shift 2)

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Series LCR Circuit

The a.c. source is connected to series LCR circuit. If voltage across $R$ is $40\text{ V}$ , that across $L$ is $80\text{ V}$ and that across $C$ is $40\text{ V}$ , then the e.m.f. ' $e$ ' of a.c. source is

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Answer: (b)

$e = \sqrt{V_R^2 + (V_L - V_C)^2}$
$e = \sqrt{40^2 + (80 - 40)^2} = \sqrt{3200} = 40\sqrt{2}$ V
So the answer is (b) $40\sqrt{2}$ V.

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Series LCR Circuit question from MHT CET 2023 (11 May, Shift 2)

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