Source: High school physics (Chinese)
Problem Sets:
Problem
Two incandescent bulbs have rated values "$220 \ \text{V}, 100 \ \text{W}$" (bulb $A$) and "$220 \ \text{V}, 60 \ \text{W}$" (bulb $B$). Their voltage--current characteristic curves are given; both curves rise monotonically and bend over (the resistance grows with voltage because the filament heats up).
Representative readings from the curves:
| $U \ (\text{V})$ | 30 | 50 | 100 | 110 | 160 | 200 |
|---|---|---|---|---|---|---|
| $I_A \ (\text{A})$ | 0.16 | 0.22 | 0.32 | 0.34 | 0.41 | 0.46 |
| $I_B \ (\text{A})$ | 0.09 | 0.13 | 0.19 | 0.20 | 0.24 | 0.28 |
- Connect the two bulbs in parallel across a $110 \ \text{V}$ source. Use the curves to find the actual power dissipated in each bulb.
- Connect the two bulbs in series across a $220 \ \text{V}$ source. Use the curves to find the actual power in each bulb, then compare with the values obtained by assuming the bulb resistances stay constant at their rated (hot) values.
- Find the resistance of the $100 \ \text{W}$ bulb at $U = 30, 50, 100, 160, 200 \ \text{V}$, and sketch the resistance-versus-voltage curve.
(1) $P_A \approx 37 \ \text{W}$, $P_B \approx 22 \ \text{W}$. (2) From the curves: $P_A \approx 15 \ \text{W}$, $P_B \approx 40 \ \text{W}$ (current $\approx 0.25 \ \text{A}$). With constant rated $R$: $P_A \approx 14.1 \ \text{W}$, $P_B \approx 23.4 \ \text{W}$ -- constant-$R$ severely underestimates $P_B$ because the cooler filaments have lower-than-rated resistance. (3) $R$ at $30, 50, 100, 160, 200 \ \text{V}$ is approximately $188, 227, 313, 390, 435 \ \Omega$; $R$ increases with $U$.
\textbf{(1) Parallel across 110 V.} Each bulb has $110 \ \text{V}$ across it. From the table $I_A(110 \ \text{V}) \approx 0.34 \ \text{A}$ and $I_B(110 \ \text{V}) \approx 0.20 \ \text{A}$:
$$P_A = U I_A = 110 \times 0.34 \approx 37 \ \text{W}, \qquad P_B = U I_B = 110 \times 0.20 = 22 \ \text{W}.$$\textbf{(2) Series across 220 V (graphical method).} The same current $I$ flows through both bulbs, and the voltages add to $220 \ \text{V}$: $U_A(I) + U_B(I) = 220 \ \text{V}$. Reading the curves, this is satisfied at $I \approx 0.25 \ \text{A}$ with $U_A \approx 60 \ \text{V}$ and $U_B \approx 160 \ \text{V}$. Hence
$$P_A \approx U_A I = 60 \times 0.25 = 15 \ \text{W}, \qquad P_B \approx U_B I = 160 \times 0.25 = 40 \ \text{W}.$$\textbf{Series with constant rated resistances.} Using $R = U_\text{rated}^2 / P_\text{rated}$:
$$R_A = \dfrac{220^2}{100} = 484 \ \Omega, \quad R_B = \dfrac{220^2}{60} \approx 806.7 \ \Omega, \quad I' = \dfrac{220}{R_A + R_B} \approx 0.170 \ \text{A}.$$ $$P_A' = I'^{\,2} R_A \approx 14.1 \ \text{W}, \qquad P_B' = I'^{\,2} R_B \approx 23.4 \ \text{W}.$$\emph{Comparison:} the constant-$R$ calculation badly underestimates the actual power in the dimmer bulb $B$ ($23.4 \ \text{W}$ vs $\sim 40 \ \text{W}$ from the curves) and slightly underestimates $P_A$. The reason: at the series current ($\approx 0.25 \ \text{A}$, well below bulb $A$'s rated $0.455 \ \text{A}$ but close to bulb $B$'s rated $0.273 \ \text{A}$), bulb $A$ is much cooler than its rated state and its filament resistance is much lower than $484 \ \Omega$, while bulb $B$ is close to its rated point and its resistance is near $806.7 \ \Omega$. So nearly all of the $220 \ \text{V}$ falls across $B$.
\textbf{(3) Resistance of bulb $A$ versus voltage.} $R(U) = U / I_A(U)$:
| $U \ (\text{V})$ | 30 | 50 | 100 | 160 | 200 |
|---|---|---|---|---|---|
| $I_A \ (\text{A})$ | 0.16 | 0.22 | 0.32 | 0.41 | 0.46 |
| $R \ (\Omega)$ | $\approx 188$ | $\approx 227$ | $\approx 313$ | $\approx 390$ | $\approx 435$ |
The $R(U)$ curve rises monotonically and tends to flatten out at higher voltages -- the filament's resistance grows with temperature, which grows with the power dissipated.