As given in the figure- a series circuit connected across a 200V - 60 Hz line consists of a capacitor of capacitive reactance 30 - a non-inductive resistor of 44- and a coil of inductive reactance 90 - and resistance 36 - The power dissipated in the coil is

The correct option is C 144 W q=vz =200L0(36+j90)+44 (j10) =200L080+j60 =200L0100 lt;37 power in 001'1=I× R =(2)^2× 36 =144 ...

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Standard XII

Physics

V-I Characteristics for a P-N Junction Diode

As given in t...

Question

As given in the figure, a series circuit connected across a $200$ V $, 60$ Hz line consists of a capacitor of capacitive reactance $30 Ω$ , a non-inductive resistor of $44 Ω$ , and a coil of inductive reactance $90 Ω$ and resistance $36 Ω$ . The power dissipated in the coil is

320 W

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176 W

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144 W

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0 W

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Solution

The correct option is C 144 W
$q = \frac{v}{z}$
$= \frac{200 L 0}{(36 + j 90) + 44 - (j 10)}$
$= \frac{200 L 0}{80 + j 60}$
$= \frac{200 L 0}{100 < 37}$
power in $001^{'} 1 = I \times R$
$= (2)^{2} \times 36$
$= 144 W$

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As given in the figure, a series circuit connected across a 200V ,60 Hz line consists of a capacitor of capacitive reactance 30Ω, a non-inductive resistor of 44Ω, and a coil of inductive reactance 90Ω and resistance 36Ω. The power dissipated in the coil is

The correct option is C 144 Wq=vz=200L0(36+j90)+44−(j10)=200L080+j60=200L0100<37power in 001′1=I×R=(2)2×36=144W

As given in the figure, a series circuit connected across a $200$ V $, 60$ Hz line consists of a capacitor of capacitive reactance $30 Ω$ , a non-inductive resistor of $44 Ω$ , and a coil of inductive reactance $90 Ω$ and resistance $36 Ω$ . The power dissipated in the coil is

The correct option is C 144 W
$q = \frac{v}{z}$
$= \frac{200 L 0}{(36 + j 90) + 44 - (j 10)}$
$= \frac{200 L 0}{80 + j 60}$
$= \frac{200 L 0}{100 < 37}$
power in $001^{'} 1 = I \times R$
$= (2)^{2} \times 36$
$= 144 W$