In the series L-C-R circuit shown,the impedan | vedclass

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

Question

(D) Given: Inductance $L = 1 	ext{ H}$,Capacitance $C = 20 \mu	ext{F} = 20 	imes 10^{-6} 	ext{ F}$,Resistance $R = 300 \Omega$,Frequency $f = \fra

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) Given: Inductance $L = 1 	ext{ H}$,Capacitance $C = 20 \mu	ext{F} = 20 	imes 10^{-6} 	ext{ F}$,Resistance $R = 300 \Omega$,Frequency $f = \frac{50}{\pi} 	ext{ Hz}$.First,calculate the inductive reactance $X_{L}$:$X_{L} = 2 \pi f L = 2 \pi \left(\frac{50}{\pi}ight) 	imes 1 = 100 \Omega$.Next,calculate the capacitive reactance $X_{C}$:$X_{C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \left(\frac{50}{\pi}ight) 	imes 20 	imes 10^{-6}} = \frac{1}{100 	imes 20 	imes 10^{-6}} = \frac{1}{2 	imes 10^{-3}} = 500 \Omega$.The impedance $Z$ of a series $L-C-R$ circuit is given by:$Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$.Substituting the values:$Z = \sqrt{(300)^{2} + (500 - 100)^{2}} = \sqrt{300^{2} + 400^{2}} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \Omega$.

In the series $L-C-R$ circuit shown,the impedance is (in $Omega$)

Similar Questions

In a series $LCR$ circuit,the voltages across $R, L$ and $C$ are shown in the figure. The voltage of the applied source is ........ $V$. (in $, V$)

The value of alternating e.m.f. $(E)$ in the given circuit is (in $V$)

In the given circuit,the magnitudes of $V_{L}$ and $V_{C}$ are twice that of $V_{R}$. Given that $f=50\,Hz$ and $R=5\,\Omega$,the inductance of the coil is $\frac{1}{K\pi}\,mH$. The value of $K$ is:

When a capacitor is connected in series with an $LR$ circuit,the alternating current flowing in the circuit

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