Question 694: Current Electricity

Question

Three electric bulbs of 200 W and 400 W are shown in figure. The resultant power of the combination, if rated voltage is applied across the combination is
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1mo60jga8/e542ff03-fef8-416f-a60f-a73cc4249c3c/698aa300-3c11-11f1-96b8-6f33e2dc7853/file-6y3zli1mo60jga9.png?format=png)

Accepted Answer

Let the resistance of the 400 W bulb be $R$. Since power is inversely proportional to resistance for the same voltage, the resistance of the 200 W bulb will be $2R$.
We know that power is given by
$ P = \frac{V^2}{R} $
So for the 400 W bulb,
$ R = \frac{V^2}{400} $
The circuit shows both bulbs connected in series. The total (net) resistance in series is the sum of their resistances:
$ R_{	ext{total}} = 2R $
The net power of the combination is then given by
$ P_{	ext{net}} = \frac{V^2}{R_{	ext{total}}} = \frac{V^2}{2R} $
Substituting $R = \frac{V^2}{400}$, we get
$ P_{	ext{net}} = \frac{V^2}{2 	imes \frac{V^2}{400}} = 200~\mathrm{W} $
Hence, the total power of the combination is
200 W
.

Answer

800 W

Answer

400 W

Answer

200 W

Answer

600 W

Current Electricity

Three electric bulbs of 200 W and 400 W are shown in figure. The resultant power of the combination, if rated voltage is applied across the combination is

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