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  • A heater is designed to operate with a power of 1000 watts in a 100 volt line. It is connected in combination with a resistance ,to a 10

A heater is designed to operate with a power of 1000 watts in a 100 volt line. It is connected in combination with a resistance \mathrm{R},to a 100 volt mains as shown in figure. The value of \mathrm{R}so that the heater may operate with a power of 62.5 watts is

Option: 1

10 \Omega


Option: 2

20 \Omega


Option: 3

30 \Omega


Option: 4

5 \Omega


Answers (1)

best_answer

The resistance of heater
\mathrm{R}_{\mathrm{H}}=\frac{\mathrm{v}_{\mathrm{H}}^{2}}{\mathrm{~W}}=\frac{100 \times 100}{1000}=10 \Omega

And as it dissipates \mathrm{62.5 \mathrm{w}}
\mathrm{\frac{V_{H}^{2}}{R_{H}}=62.5 \text { i.e. } V_{H}^{2}=62.5 \times 10 \text {. It gives } V_{H}=25 \mathrm{~V}}

Now as applied voltage is \mathrm{100 \mathrm{V}},
\mathrm{100=V_{H}+V_{10} \text {, i.e. } V_{10}=100-25=75 V}

and hence circuit current
\mathrm{\mathrm{I}=\mathrm{I}_{10}=\frac{\mathrm{V}_{10}}{\mathrm{R}_{10}}=\frac{75}{10}=7.5 \mathrm{~A}}

so if \mathrm{R}is the unknown resistance
\mathrm{I}=\mathrm{I}_{\mathrm{H}}+\mathrm{I}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{H}}}{\mathrm{R}_{\mathrm{H}}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}}

\mathrm{But\, \mathrm{I}=7.5 \mathrm{~A}, \mathrm{~V}_{\mathrm{H}}=\mathrm{V}_{\mathrm{R}}=25 \mathrm{~V}, \mathrm{R}_{\mathrm{H}}=10 \Omega}
\mathrm{7.5=\frac{25}{10}+\frac{25}{R}}

Hence \mathrm{\mathrm{R}=5 \Omega}.

Posted by

Shailly goel

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