Question.
Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.
Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.
solution:
Here, charge, $\mathrm{Q}=96000 \mathrm{C} ;$ time, $\mathrm{t}=1 \mathrm{hr}$
potential difference, $\mathrm{V}=50 \mathrm{~V}$.
Heat produced, $\mathrm{H}=\mathrm{V} \mathrm{I} \mathrm{t}=\mathrm{V} \times \mathrm{Q} \quad[\because \mathrm{Q}=\mathrm{I} \mathrm{t}]$
$=96000 \mathrm{C} \times 50 \mathrm{~V}=4.8 \times 10^{6} \mathrm{~J}$
Here, charge, $\mathrm{Q}=96000 \mathrm{C} ;$ time, $\mathrm{t}=1 \mathrm{hr}$
potential difference, $\mathrm{V}=50 \mathrm{~V}$.
Heat produced, $\mathrm{H}=\mathrm{V} \mathrm{I} \mathrm{t}=\mathrm{V} \times \mathrm{Q} \quad[\because \mathrm{Q}=\mathrm{I} \mathrm{t}]$
$=96000 \mathrm{C} \times 50 \mathrm{~V}=4.8 \times 10^{6} \mathrm{~J}$
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