Q. calculate the value of the resistor to be connected in series with a 100W;
230V lamp so that the wattage of the lamp is reduced to 35W
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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
| Is This Answer Correct ? | 3 Yes | 1 No |
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.
| Is This Answer Correct ? | 2 Yes | 0 No |
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
| Is This Answer Correct ? | 1 Yes | 0 No |
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
| Is This Answer Correct ? | 0 Yes | 0 No |
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