Electric Circuits
Intermediate
Ohm's Law
Source: High school physics (Chinese)
Problem
A potentiometer is labeled $1 k\Omega$, $40 W$.
- What is the maximum allowable current through this potentiometer?
- What is the maximum allowable voltage across this potentiometer?
- When $10 V$ is applied across this potentiometer, what is its electrical power dissipation?
(1) $I_{\max} = 0.2 A$; (2) $U_{\max} = 200 V$; (3) $P = 0.1 W$.
Let $R = 1 k\Omega = 1000 \Omega$ and $P_{\max} = 40 W$ be the rated values.
(1) From $P_{\max} = I_{\max}^2 R$:
$$I_{\max} = \sqrt{P_{\max} / R} = \sqrt{40 / 1000} = 0.2 A.$$(2) From $P_{\max} = U_{\max}^2 / R$:
$$U_{\max} = \sqrt{P_{\max} \, R} = \sqrt{40 \times 1000} = 200 V.$$(3) At applied voltage $U = 10 V$ (well within the rated maximum), the dissipated power is:
$$P = U^2 / R = 10^2 / 1000 = 0.1 W.$$