In the given circuit,the magnitudes of V_{L} | vedclass

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

Question

(D) Given: $V_{L} = V_{C} = 2V_{R}$.Since $V = IR$,we have $I X_{L} = I X_{C} = 2(IR)$.Therefore,$X_{L} = X_{C} = 2R$.Given $R = 5\,\Omega$,so $X_{L}

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(D) Given: $V_{L} = V_{C} = 2V_{R}$.Since $V = IR$,we have $I X_{L} = I X_{C} = 2(IR)$.Therefore,$X_{L} = X_{C} = 2R$.Given $R = 5\,\Omega$,so $X_{L} = 2 	imes 5 = 10\,\Omega$.We know $X_{L} = 2\pi f L$.Substituting the values: $10 = 2 	imes \pi 	imes 50 	imes L$.$10 = 100 \pi L$.$L = \frac{10}{100\pi} = \frac{1}{10\pi}\,H$.To convert to $mH$,$L = \frac{1}{10\pi} 	imes 1000\,mH = \frac{100}{\pi}\,mH$.Comparing this with $\frac{1}{K\pi}\,mH$,we have $\frac{1}{K\pi} = \frac{100}{\pi}$.$K = \frac{1}{100} = 0.01$.

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:

Similar Questions

In the adjoining figure,the impedance of the circuit will be $... \Omega$.

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

The average power consumed by the given circuit is $W$.

In the $RLC$ series circuit as shown in the diagram,the maximum value of charge on the capacitor is:

In a series $RLC$ circuit,the $r.m.s.$ voltages across the resistor and the inductor are respectively $400 \,V$ and $700 \,V$. If the equation for the applied voltage is $\varepsilon = 500 \sqrt{2} \sin \omega t$,then the peak voltage across the capacitor is ........... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module