Question

a wire of 15 ohm resistance is gradually stretched to twice its length.It is then cut into two equal parts.these parts are then connected in parallel across a 3.0V battery.Find the current drawn from he battery.

Answer 1

R'=n²R=\((2)^2\times15=60\Omega\)

Resistance of each half part=\(\frac{60}{2}=30\Omega\)

When both parts are connected in parallel, the final resistance=\(\frac{30}{2}=15\Omega\)

Current drawn from the battery, I=\(\frac{V}{R}\)=\(\frac{0.3}{15}\)=0.2A