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

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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