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