As shown in the figure,a series circuit conne | vedclass

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

Question

(C) The total resistance of the circuit is $R = R_1 + R_2 = 44\,\Omega + 36\,\Omega = 80\,\Omega$.The net reactance of the circuit is $X = X_L - X_C =

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(C) The total resistance of the circuit is $R = R_1 + R_2 = 44\,\Omega + 36\,\Omega = 80\,\Omega$.The net reactance of the circuit is $X = X_L - X_C = 90\,\Omega - 30\,\Omega = 60\,\Omega$.The impedance of the circuit is $Z = \sqrt{R^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$.The current in the series circuit is $I = \frac{V}{Z} = \frac{200\, V}{100\,\Omega} = 2\, A$.The power dissipated in the coil is due to its internal resistance $R_2 = 36\,\Omega$. Therefore,$P = I^2 R_2 = (2)^2 	imes 36 = 4 	imes 36 = 144\, W$.

Similar Questions

The reading of the ammeter in the circuit is ............ $A$.

$A$ $50 \,W$ bulb connected in series with a heater coil is put to an $AC$ mains. Now the bulb is replaced by a $100 \,W$ bulb. The heater output will ...........

An inductor of reactance $100 \ \Omega$, a capacitor of reactance $50 \ \Omega$, and a resistor of resistance $50 \ \Omega$ are connected in series with an $AC$ source of $10 \ V, 50 \ Hz$. Average power dissipated by the circuit is . . . . . . $W$.

The figure shows the variation of $R$,$X_L$,and $X_C$ with frequency $f$ in a series $L-C-R$ circuit. For what frequency point is the circuit capacitive?

$A$ series $LCR$ circuit containing a resistance of $120 \, \Omega$ has an angular resonance frequency of $4 \times 10^3 \, rad \, s^{-1}.$ At resonance, the voltages across the resistance and the inductance are $60 \, V$ and $40 \, V$ respectively. The values of $L$ and $C$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module