An inductor of inductance L,a capacitor of ca | vedclass

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

Question

(D) Given:Amplitude of current,$I_{0} = 10 \sqrt{2} \, A$.Potential difference across inductor,$V_{L} = 40 \, V$.Potential difference across capacitor

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Given:Amplitude of current,$I_{0} = 10 \sqrt{2} \, A$.Potential difference across inductor,$V_{L} = 40 \, V$.Potential difference across capacitor,$V_{C} = 10 \, V$.Potential difference across resistor,$V_{R} = 40 \, V$.Step $1$: Calculate the $RMS$ current $(I_{RMS})$.$I_{RMS} = \frac{I_{0}}{\sqrt{2}} = \frac{10 \sqrt{2}}{\sqrt{2}} = 10 \, A$.Step $2$: Calculate the $RMS$ voltage of the source $(V_{RMS})$.In an $LCR$ series circuit,the total voltage is given by:$V_{RMS} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$$V_{RMS} = \sqrt{(40)^{2} + (40 - 10)^{2}}$$V_{RMS} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, V$.Step $3$: Calculate the impedance $(Z)$ of the circuit.$Z = \frac{V_{RMS}}{I_{RMS}} = \frac{50 \, V}{10 \, A} = 5 \, \Omega$.Therefore,the impedance of the circuit is $5 \, \Omega$.

Similar Questions

In the circuit diagram shown,$X_C = 100 \, \Omega$,$X_L = 200 \, \Omega$,and $R = 100 \, \Omega$. The source voltage is $V = 200 \sin(\omega t) \, \text{V}$. The effective $(RMS)$ current through the source is:

If a resistance of $100\, \Omega$,an inductance of $0.5\, H$,and a capacitance of $10 \times 10^{-6}\, F$ are connected in series to a $50\, Hz$ $AC$ supply,then the impedance is:

In an $R-L-C$ series circuit, the potential difference across each element is $20 \, V$. If the value of the resistance $R$ is doubled, what will be the potential difference across $R, L$, and $C$ respectively?

In a series $LCR$ circuit,$R$ represents the resistance of an electric bulb. If the frequency of the $A.C.$ supply is doubled,what should be the values of inductance $L$ and capacitance $C$ to maintain the same current?

Power dissipated in an $LCR$ series circuit connected to an $a.c.$ source of electromotive force (emf) $\varepsilon$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module