Electricity And Magnitism

Energy consumed by an appliance is given by the expression,

$P=\frac { V }{ I }$

$R=\frac { { V }^{ 2 } }{ P }$

Where,

Power rating, $P = 100 W$

Voltage, $V = 220 V$

Resistance, $R = \frac { { 220 }^{ 2 } }{ 100 } =484\Omega$

The resistance of the bulb remains constant if the supply voltage is reduced to $110 V$ . If the

bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as

$\therefore P=\frac { ({ V }^{ 2 }) }{ R } =\frac { { 110 }^{ 2 } }{ 484 } =25W$

Therefore, the power consumed will be $25 W$

Power, P = I*V = I^2 * R = V^2/R

with respect to above equations power consumed is not 25W

any thoughts???

yes! that's ri8 about P=VI we know that electric bulb=220V and Power consumed=100W so 1st we determined the current flowing in the circuit which I=100W/220V=0.454545 Now the condition is that the when we decreased the the Voltage 110V, So again we put P=IV=0.4545x110=50W if we compare the intial and final voltages is 220/110=2 that why 50W/2=25W

P=V^2/R Find resistance with its help and then the power..