A 100W- 200V bulb is connected to a 160V supply- The actual power consumption would be185W100W64W54W

Question

Toppr Admin · Accepted Answer

Resistance of the bulb R-V2Pwhere P-100W and V-200 V-x2234- R-200-2100-400-x3A9-Actual power consumption P-x2032-V21Rwhere V1-160 V-x2234- P-x2032-160-2400-64W

A $100 W, 200 V$ bulb is connected to a $160 V$ supply. The actual power consumption would be
$185 W$
$100 W$
$64 W$
$54 W$

Resistance of the bulb $R = \frac{V^{2}}{P}$
where $P = 100 W$ and $V = 200$ V.
$∴$ $R = \frac{(200)^{2}}{100} = 400 Ω$
Actual power consumption $P^{'} = \frac{V_{1}^{2}}{R}$
where $V_{1} = 160$ V
$∴$ $P^{'} = \frac{(160)^{2}}{400} = 64 W$

A 100W,200V bulb is connected to a 160V supply. The actual power consumption would be185W100W64W54W

Resistance of the bulb R=V2Pwhere P=100W and V=200 V.∴ R=(200)2100=400ΩActual power consumption P′=V21Rwhere V1=160 V∴ P′=(160)2400=64W

A $100 W, 200 V$ bulb is connected to a $160 V$ supply. The actual power consumption would be
$185 W$
$100 W$
$64 W$
$54 W$

Resistance of the bulb $R = \frac{V^{2}}{P}$
where $P = 100 W$ and $V = 200$ V.
$∴$ $R = \frac{(200)^{2}}{100} = 400 Ω$
Actual power consumption $P^{'} = \frac{V_{1}^{2}}{R}$
where $V_{1} = 160$ V
$∴$ $P^{'} = \frac{(160)^{2}}{400} = 64 W$