Source: High school physics (Chinese)
Problem Sets:
Problem
Four identical light bulbs, each rated at $U_0 = 12$ V and $P_0 = 24$ W, are connected in parallel and powered by a source with EMF $\varepsilon = 12$ V and internal resistance $r = 0.20\ \Omega$. Assume the filament resistance is constant (independent of current).
- When only one bulb is turned on, what is the voltage across the bulb?
- When all four bulbs are turned on, what is the voltage across each bulb?
(1) $U_1 \approx 11.61$ V.\ (2) $U_4 \approx 10.59$ V.
Each bulb at rated conditions: $I_0 = P_0/U_0 = 24/12 = 2$ A, so the filament resistance is $R = U_0/I_0 = 6\ \Omega$.
(1) One bulb in series with $r$:
$$I_1 = \frac{\varepsilon}{R + r} = \frac{12}{6 + 0.20} = \frac{60}{31}\ \text{A}.$$Voltage across the bulb: $U_1 = I_1 R = \dfrac{360}{31} \approx 11.61$ V.
(2) Four bulbs in parallel: $R_\parallel = R/4 = 1.5\ \Omega$. Total current:
$$I_4 = \frac{\varepsilon}{R_\parallel + r} = \frac{12}{1.7} = \frac{120}{17}\ \text{A}.$$Voltage across each bulb: $U_4 = I_4 R_\parallel = \dfrac{180}{17} \approx 10.59$ V.