Q#1073: [Current of Electricity](Past Exam Paper – J85 / I / 17)
Three identical cells each having an e.m.f. of 1.5 V and a constant internal resistance of 2.0 Ω are connected in series with a 4.0 Ω resistance R, firstly as in circuit (i), and secondly as in circuit (ii).
What is the ratio power in R in circuit (i) / power in R in circuit (ii)?
A 9.0 B 7.2 C 5.4 D 3.0 E 1.8
Solution 1073:
Answer: A.
Notice that in circuit (ii), the negative terminals of 2 of the cells are connected to each other, instead of the negative terminal being connected to the positive terminal of the other cell (the same goes for the positive terminal of that middle cell).
Current flows in a specific direction, so this affects the overall e.m.f. in the circuit.
Overall e.m.f. in circuit (i) = 1.5 + 1.5 + 1.5 = 4.5 V
Overall e.m.f. in circuit (ii) = 1.5 – 1.5 + 1.5 = 1.5 V
In each case, the corresponding p.d. would be across the resistor R.
The total internal resistance due to the cells are the same in both cases. Internal resistance is due to the cells, it does not have a specific direction, so the total is the same in both cases.
Total internal resistance due to the cells = 2.0 + 2.0 + 2.0 = 6.0 Ω
Total resistance in each circuit = 6.0 + 4.0 = 10.0 Ω (same for both cases)
Power in resistor R = I2R = V2 / R
Power P1 in resistor R in circuit (i) = 4.52 / 4.0 = 5.0625 W
Power P2 in resistor R in circuit (ii) = 1.52 / 4.0 = 0.5625 W
Ratio = P1 / P2 = 5.0625 / 0.5625 = 9.0
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