In a series LCR circuit, the rms voltage acro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The applied voltage is given by $V = 50 \sqrt{2} \sin \omega t$. The peak voltage is $V_0 = 50 \sqrt{2} \ V$, so the $rms$ voltage is $V_{rms} = \

Vedclass · Accepted Answer

$50 \sqrt{2} \ V$

Vedclass · Answer

(D) The applied voltage is given by $V = 50 \sqrt{2} \sin \omega t$. The peak voltage is $V_0 = 50 \sqrt{2} \ V$, so the $rms$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = 50 \ V$.In a series $LCR$ circuit, the relationship between the $rms$ voltages is given by $V_{rms}^2 = V_R^2 + (V_L - V_C)^2$.Given: $V_R = 30 \ V$, $V_C = 90 \ V$, and $V_{rms} = 50 \ V$.Substituting the values: $50^2 = 30^2 + (V_L - 90)^2$.$2500 = 900 + (V_L - 90)^2$.$(V_L - 90)^2 = 1600$.$V_L - 90 = \pm 40$.Case $1$: $V_L = 90 + 40 = 130 \ V$.Case $2$: $V_L = 90 - 40 = 50 \ V$.Assuming the standard case where the inductor voltage is the peak value asked, we look for the peak voltage across the inductor. The $rms$ voltage $V_L$ is $50 \ V$ or $130 \ V$. The peak voltage is $(V_L)_{peak} = V_L \sqrt{2}$.For $V_L = 50 \ V$, $(V_L)_{peak} = 50 \sqrt{2} \ V$.

In a series $LCR$ circuit, the $rms$ voltage across the resistor and the capacitor are $30 \ V$ and $90 \ V$ respectively. If the applied voltage is $V = 50 \sqrt{2} \sin \omega t$, then the peak voltage across the inductor is

Similar Questions

$A$ light bulb connected in series with a capacitor and an $a.c.$ source is glowing with certain brightness. On reducing the value of capacitance and frequency respectively, the brightness of the bulb

An a.c. source of frequency $f$ is connected to a circuit containing an inductance $L$ and resistance $R$ in series. The impedance of this circuit is

In an $AC$ circuit,a resistance of $R$ $\Omega$ is connected in series with an inductor of self-inductance $L$. If the phase angle between voltage and current is $45^{\circ}$,the value of inductive reactance $(X_{L})$ will be equal to . . . . . . .

$A$ coil of $200 \, \Omega$ resistance and $1.0 \, H$ inductance is connected to an $ac$ source of frequency $200/2\pi \, Hz.$ The phase angle between potential and current will be.....$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module