In the given circuit of below the cell P has an e.m.f. 1.5V and an unknown internal resistance r while the cell Q has an e.m.f. 2.0V and an internal resistance of 1ohm. If the ammeter reads 50mA, then r is equal to A. 0.5 ohm B.0.62 ohm C. 1.0 ohm D. 2.0 ohms E. 3.0 ohms.
ANSWER: Option D
EXPLANATION: Since the cells are connected -ve to -ve, the total emf will be 2.0V – 1.5V = 0.5V
emf = I(R + r + r2)
Where; emf is the resultant Volts the two cells gives
I is the current in the circuit
r is the internal resistance of the first cell
r2 is the internal resistance of the second cell
0.5 = 50 × 10⁻³ (7 + r + 1)
0.5 = 50 × 10⁻³ (8 + r)
Now, solve for r to get your answer!
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His Name is Tiamiyu Abdulbazeet Olawale. He is a student at the University of Ilorin studying Electrical and Electronics Engineering. He likes to give the latest updates about Nigeria and international institutions as he aims at helping people to lay their hands on the right information.