An inductor 20 mH- a capacitor 100 muF and a resistor 50 Omega are connected in series across a of emf- V-10 sin 314t- The power loss in the circuit is-2-74 W1-13 W0-79 W0-43 W

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Toppr Admin · Accepted Answer

L-20mH C-100-x3BC-F R-50-x3A9-V-10sin-314t-V0-10- -x3C9-314XL-wL-314-xD7-20-xD7-10-x2212-3-6-28-x3A9-XC-1-x3C9-C-31-8-x3A9-Z-x221A-R2-XC-x2212-XL-2-56-1Power-xA0-lossP-V20R2Z2-0-79W

An inductor $20$ mH, a capacitor $100$ $μ$ F and a resistor $50$ $Ω$ are connected in series across a source of emf, V $= 10$ $sin 314$ t. The power loss in the circuit is?
$2.74$ W
$1.13$ W
$0.79$ W
$0.43$ W

$L = 20 m H$ $C = 100 μ F$ $R = 50 Ω$
$V = 10 s i n (314 t)$
$V_{0} = 10$ , $ω = 314$
$X_{L} = w L = 314 \times 20 \times 10^{- 3} = 6.28 Ω$
$X_{C} = \frac{1}{ω C} = 31.8 Ω$
$Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}} = 56.1$
$P o w e r l o s s P = \frac{V_{0}^{2} R}{2 Z^{2}} = 0.79 W$

An inductor 20 mH, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V=10 sin314t. The power loss in the circuit is?2.74 W1.13 W0.79 W0.43 W

L=20mH C=100μF R=50ΩV=10sin(314t)V0=10, ω=314XL=wL=314×20×10−3=6.28ΩXC=1ωC=31.8ΩZ=√R2+(XC−XL)2=56.1Power lossP=V20R2Z2=0.79W

An inductor $20$ mH, a capacitor $100$ $μ$ F and a resistor $50$ $Ω$ are connected in series across a source of emf, V $= 10$ $sin 314$ t. The power loss in the circuit is?
$2.74$ W
$1.13$ W
$0.79$ W
$0.43$ W

$L = 20 m H$ $C = 100 μ F$ $R = 50 Ω$
$V = 10 s i n (314 t)$
$V_{0} = 10$ , $ω = 314$
$X_{L} = w L = 314 \times 20 \times 10^{- 3} = 6.28 Ω$
$X_{C} = \frac{1}{ω C} = 31.8 Ω$
$Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}} = 56.1$
$P o w e r l o s s P = \frac{V_{0}^{2} R}{2 Z^{2}} = 0.79 W$