Question

If a conductor with a current of 200 mA passing through it converts 40 J of electrical energy into heat in 30 s, what is the potential drop across the conductor?

Answer 1

We use the relationship $P=\frac{\Delta E}{\Delta t}= I \Delta V$ to find the potential drop ∆V:

$\Delta V= \frac{\Delta E}{I\Delta t}= \frac{40\,J}{(0.200 A)(30\,s)}= \frac{20}{3}V \approxeq 6.67\,V$

In conclusion, the potential drop across the conductor is ∆V=6.67 V.