Answer / 230

power = 100 W,= VI
volts = 230v,
I=V/P;
I=100/230;
=0.4347;
R=V/I;
=230/0.4347;
I=529.1,
NEXT POWER REDUCED TO 35 MEANS, 2.8* R;
=2.85*529.1;
=1507.9 Ohms.
That is 1507.9 Ohms to be increased to get 35 W

Answer / sagar

power P =VI
I = 100/230 = 0.4347A
when there is no addition resistance connectedt,then
R = V/I = 230/0.4347 =529 ohm.


If we want to reduce power to 35 w ,then
I = 35/230 = 0.1521A should be flow
for that value of resistance
R = V/I = 230/0.1521
=1512 ohm.

so,series resistance conneced will be,
= 1512 - 529
= 982 ohm.

Answer / gmpk

dear friend,where is the series resistor in the circuit.

the actual question that what is the resistance between

source and load if the load is 35w instead of 100w.

100/35 is 2.8 times has reduced. so if we multiply the actual

value of resistance will come

Answer / michael

982.4285 ohms

Answer / kalulu

lamp 100W/230v
when resistor R is connected in series with the lamp.
lamp power=35W
lamp Voltage= 230V /* since 230v is the operating voltage of the lamp*/
series current = 35/230
=0.152 A
power dissipated at a resistor = 100 W -35 W
= 65 W
= Isqr X R
R= 65/(0.152sqr)
= 2.8 kilo-ohm

Answer / lakshmikandhan.v

813.8 ohm

Answer / anand pathak

813.8

Answer / ashok naik

64.5w

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