In the series L-C-R circuit shown in the figure, the rms voltage across the resistor and inductor are 400 V and 700 V respectively. If the applied voltage is E=500√2 sin(ω t) , then the peak voltage across the capacitor is .

Question

In the series L-C-R circuit shown in the figure, the rms voltage across the resistor and inductor are 400 V and 700 V respectively. If the applied voltage is E=500√2 sin(ω t) , then the peak voltage across the capacitor is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pspzaovugcyb.png" /> . (A) 1200 V (B) 1200√2V (C) 400 V (D) 400√2V

Tardigrade Admin · Accepted Answer

VR2+(VL-VC2=Er m s2 (VL - VC)2=(500)2 - (400)2 VL - VC=√( 5002 - ( 400 2=300 VC=VL - 300=700 - 300=400 ∴ ( textV) textC ( textpeak) = √2 ( textV) textc = √2 × 4 0 0 textvolts

Q. In the series $L - C - R$ circuit shown in the figure, the rms voltage across the resistor and inductor are $400 V$ and $700 V$ respectively. If the applied voltage is $E = 5002 s in (ω t)$ , then the peak voltage across the capacitor is

.

$1200 V$

$12002 V$

$400 V$

$4002 V$

$V_{R}^{2} + (V_{L} - V_{C}^{2} = E_{r m s}^{2}$
$(V_{L} - V_{C})^{2} = (500)^{2} - (400)^{2}$
$V_{L} - V_{C} = (50 0^{2} - (40 0^{2} = 300$
$V_{C} = V_{L} - 300 = 700 - 300 = 400$
$∴ (V)_{C} (peak) = 2 (V)_{c} = 2 \times 400 volts$

Q. In the series L−C−R circuit shown in the figure, the rms voltage across the resistor and inductor are 400V and 700V respectively. If the applied voltage is E=5002​sin(ωt) , then the peak voltage across the capacitor is .

1200V

12002​V

400V

4002​V