Source: High school physics (Chinese)
Problem
A power source with EMF $30 V$ and internal resistance $1 \Omega$ is connected in a closed series loop with a light bulb rated $6 V$, $12 W$ and a motor whose coil resistance is $2 \Omega$. The bulb operates exactly at its rated values.
- What is the heat-dissipation power of the motor?
- What is the mechanical (external-work) output power of the motor?
Heat-dissipation power of the motor: $P_{\text{heat}} = 8 W$. Mechanical output power: $P_{\text{mech}} = 36 W$.
The bulb at its rated values carries $I = P_L / U_L = 12 / 6 = 2 A$, which is the series-loop current.
Voltage across the motor: $U_M = \varepsilon - U_L - I r = 30 - 6 - 2 \times 1 = 22 V$.
Heat-dissipation power of the motor (resistive loss in the coil):
$$P_{\text{heat}} = I^2 R_M = 2^2 \times 2 = 8 W.$$Total electrical power input to the motor: $P_{\text{in}} = U_M I = 22 \times 2 = 44 W$.
Mechanical output power (work done externally): $P_{\text{mech}} = P_{\text{in}} - P_{\text{heat}} = 44 - 8 = 36 W$.