Two bulbs marked 200 watt - 250 volts and 100 watt - 250 volts are joined in series to 250 watt supply.?

Question

Then consumed power in the circuit is ?

Answer 1

I think you meant that they were connected in series to a 250 Volt supply. Supply is not normally rated in Watts.

OK, now let us proceed.

250 V, 200 watts means that the current is 0.8 amps (let us use DC now so that we don't have to deal with power factors and RMS / Peak voltages etc.). So, the resistance will be 250 / 0.8 = 312.5 ohms.

Remember, V X I = power and V / I = Resistance where V is voltage and I is the current.

Now the second bulb is rated as 250 v and 100 watts so current will be 0.4 amps and the resistance will be 250 / 0.4 = 625 ohms.

When both the bulbs are connected in series, the total resistance will be the sum of the two resistances, 625 + 312.5 = 937.5 ohms and the current drawn will be
250 / 937.5 = 0.267 amps and the power in the circuit will be 250 X 250 / 937 .5 = 66.67 watts

Answer 2

Cannot be determined from the information given, because the resistance of light bulbs is a non-linear function of the applied voltage. Presumably you meant 250 volt supply, rather than 250 watt. If the lamps were ideal resistors, then the 100 watt bulb would be running at 166 volts, dissipating 4/9 of 100 watts; the 200 watt bulb would be running at 83 volts, dissipating 1/9 of 200 watts.

Answer 3

200 Watt Bulbs

Answer 4

The 100 watt bulb will burn out as soon as power is turned on.
The 200 watt bulb will not stay lit, because they are in series.

Answer 5

The Amp hour score is a lengthy-time period measurement, amps are a right away fee. So that is in all likelihood that your batteries can positioned out more effective than that for a couple of minutes, yet received't keep up more effective than 7 amp-hours. So do not anticipate to run your moped at complete % for extremely lengthy. (easily accelerating and uphills take the total output, protecting a particular %, see you later because that isn't any longer the acceptable %, makes use of a lot a lot less potential).

Answer 6

We know that P = VI = V^2/R

r1 = 312.5 Q (Ohms)
r2 = 625 Q

R = r1+r2 = 937.5 Q

Consumed power in the circuit is same as current drawn is less. = 250 Watts.

Answer 7

@Tanzeeb as far as we are discussing it the0raticaIIy ....... the answer given ab0ve is c0rrect with s0me excepti0ns incIuding the 0ne y0u have 0utIined.....but in reaIity resistance 0f the Iight buIb is a functi0n 0f temperature 0f fiIament.....and it has duaI pr0p0rti0n aIs0 if we taIk ab0ut p0wer c0nsumpti0n i.e I^2 x R and V^2/R .......... s0000 there is reaIIy n0 need t0 discuss that f00Iish statement

Answer 8

Ok, get back to us when they set you free from the mental institution. 250 W of cause!