Source: High school physics (Chinese)
Problem Sets:
Problem
Three identical resistors are first connected in series to a power source, then connected in parallel to the same power source. The internal resistance of the power source is negligible. For the two connection methods, find the ratios (series to parallel) of:
- The current through each resistor.
- The total resistance of the circuit.
- The total current of the circuit.
- The power dissipated by each resistor.
- The total power dissipated by the circuit.
(1) Current through each resistor: $I_{\text{series}} : I_{\text{parallel}} = 1 : 3$
(2) Total resistance: $R_s : R_p = 9 : 1$
(3) Total current: $I_s : I_p = 1 : 9$
(4) Power per resistor: $P_{1s} : P_{1p} = 1 : 9$
(5) Total power: $P_s : P_p = 1 : 9$
Let $R$ be the resistance of each resistor and $U$ the EMF (terminal voltage, since internal resistance is zero).
\textbf{Series connection:} Total resistance $R_s = 3R$. The total current $I_s = \dfrac{U}{3R}$ also passes through each resistor. Power per resistor $P_{1s} = I_s^2 R = \dfrac{U^2}{9R}$. Total power $P_s = 3 P_{1s} = \dfrac{U^2}{3R}$.
\textbf{Parallel connection:} Total resistance $R_p = \dfrac{R}{3}$. Voltage across each resistor equals $U$, so current through each $I_{1p} = \dfrac{U}{R}$, and total current $I_p = 3 I_{1p} = \dfrac{3U}{R}$. Power per resistor $P_{1p} = \dfrac{U^2}{R}$, total power $P_p = 3 P_{1p} = \dfrac{3U^2}{R}$.
Forming the ratios (series : parallel):
$$\text{(1) Current per resistor: } \dfrac{U/(3R)}{U/R} = \dfrac{1}{3}$$ $$\text{(2) Total resistance: } \dfrac{3R}{R/3} = \dfrac{9}{1}$$ $$\text{(3) Total current: } \dfrac{U/(3R)}{3U/R} = \dfrac{1}{9}$$ $$\text{(4) Power per resistor: } \dfrac{U^2/(9R)}{U^2/R} = \dfrac{1}{9}$$ $$\text{(5) Total power: } \dfrac{U^2/(3R)}{3U^2/R} = \dfrac{1}{9}$$