Question 316: Current Electricity

Question

A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be :

Accepted Answer

Since rate of heat $$ = {{{V^2}} \over R} \Rightarrow $$ rate of heat $$ \propto {1 \over R}$$, where $$R = {{\rho L} \over A} = {{\rho L} \over {\pi {r^2}}}$$; here r is radius of wire and L is length of wire. Therefore,
$$R \propto {L \over {{r^2}}}$$.
Thus, rate of heat $$ \propto {{{r^2}} \over L}$$.
When length in halved and radius is doubled then, rate of heat $$ \propto {{{{(2r)}^2}} \over {L/2}} = {{8{r^2}} \over L}$$
Therefore, rate of heat developed in wire will be increased 8 times.

Answer

Doubled

Answer

Halved

Answer

Unchanged

Answer

Increased 8 times

Current Electricity

A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be :

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