A circuit contains an inductance of frac{1}{6 | vedclass

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

Question

(C) Given: Inductance $L = \frac{1}{6 \pi} 	ext{ H}$,Resistance $R = 15 	ext{ } \Omega$,Voltage $V = 100 	ext{ V}$,Frequency $f = 60 	ext{ Hz}$.Fi

Vedclass · Accepted Answer

$4 	ext{ A}$ and $	an^{-1}\left(\frac{4}{3}ight)$

Vedclass · Answer

(C) Given: Inductance $L = \frac{1}{6 \pi} 	ext{ H}$,Resistance $R = 15 	ext{ } \Omega$,Voltage $V = 100 	ext{ V}$,Frequency $f = 60 	ext{ Hz}$.First,calculate the inductive reactance $X_L$:$X_L = L \omega = L(2 \pi f) = \left(\frac{1}{6 \pi}ight) 	imes 2 \pi 	imes 60 = 20 	ext{ } \Omega$.Next,calculate the impedance $Z$ of the $LR$ series circuit:$Z = \sqrt{R^2 + X_L^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 	ext{ } \Omega$.The current $I$ in the circuit is:$I = \frac{V}{Z} = \frac{100}{25} = 4 	ext{ A}$.The phase difference $\phi$ between voltage and current is given by:$	an \phi = \frac{X_L}{R} = \frac{20}{15} = \frac{4}{3}$.Therefore,$\phi = 	an^{-1}\left(\frac{4}{3}ight)$.

Similar Questions

The following series $L-C-R$ circuit,when driven by an emf source of angular frequency $70 \text{ krad/s}$,behaves effectively as:

In an $A.C.$ circuit,a resistance $R = 40 \ \Omega$ and an inductance $L$ are connected in series. If the phase angle between voltage and current is $45^{\circ}$,then the value of the inductive reactance is $(\tan 45^{\circ} = 1)$. (in $\Omega$)

In the following circuit,an $AC$ input $V_i(t) = (20 \text{ mV}) \sin(10^5 t)$ is applied at the left end. The amplitude of the output voltage $V_0$ at the right end across the capacitor will be (in $mV$)

$A$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module