A heater A gives out 300, W of heat when conn | vedclass

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Question

The two heaters with different power are connected in series,

$P 1=\frac{V^{2}}{R 1}$ and $P 2=\frac{V^{2}}{R 2}$

therefore,

$R 1=\frac{V^{2}

Vedclass · Accepted Answer

$200$

Vedclass · Answer

The two heaters with different power are connected in series,

$P 1=\frac{V^{2}}{R 1}$ and $P 2=\frac{V^{2}}{R 2}$

therefore,

$R 1=\frac{V^{2}}{P 1}$ and $R 2=\frac{V^{2}}{P 2}$

in series combination

$R_{e q}=R 1+R 2$

Hence,

$P_{e q}=\frac{V^{2}}{P 1}+\frac{V^{2}}{P 2}$

$P_{e q}=\left(\frac{1}{2}\right) V^{2}$

$P_{e q}=\frac{1}{2}(400)=200 W$

A heater $A$ gives out $300\, W$ of heat when connected to a $200\, V$ $d.c.$ supply. Asecond heater $B$ gives out $600\, W$ when connected to a $200\,v$ $d.c$. supply. If a series combination of the two heaters is connected to a $200\, V$ $d.c$. supply the heat output will be ................. $W$

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