NECO 2001 Physics · Question 15

A cell can supply currents of 0.60 A and 0.30 A through a 2.0 ω and a 5.0 ω resistor respectively. What is the internal resistance of the cell?

Options

A.0.2 ω
B.0.4 ω
C.1.0 ω
D.3.0 ω
E.9.0 ω

Answer: C. 1.0 ω

Explanation written with AI and checked against the answer key

Analyze the question: The same cell (fixed EMF E and internal resistance r) drives different currents through two different external resistors. We use E = I(R + r) for each case. Concept: For a cell with internal resistance r connected to an external resistor R: E = I(R + r) Solution: Case 1: I = 0.60 A, R = 2.0 Ω E = 0.60(2.0 + r) = 1.2 + 0.60r Case 2: I = 0.30 A, R = 5.0 Ω E = 0.30(5.0 + r) = 1.5 + 0.30r Since E is the same in both cases: 1.2 + 0.60r = 1.5 + 0.30r 0.60r − 0.30r = 1.5 − 1.2 0.30r = 0.3 r = 1.0 Ω Check: E = 0.60(2 + 1) = 1.8 V, and E = 0.30(5 + 1) = 1.8 V ✓ Answer: The internal resistance is 1.0 Ω, which is option C. Tip: When a question gives two different currents and resistances for the same cell, write E = I(R + r) twice and equate them. The EMF is the constant that links both equations. Always verify by plugging r back in to see that both give the same E.

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