Three cells with e.m.f.s V1, V2 and V3, have negligible internal resistance. These cells are connected to three resistors with resistances R1, R2 and R3, as shown.

Question 6

Three cells with e.m.f.s V1, V2 and V3, have negligible internal resistance. These cells are

connected to three resistors with resistances R1, R2 and R3, as shown.

The current in the circuit is I.

Which equation is correct?

A V1 + V2 + V3 = I (R1 + R2 + R3)

B V1 + V2 – V3 = I (R1 + R2 + R3)

C V1 – V2 + V3 = I (R1 + R2 + R3)

D V1 – V2 – V3 = I (R1 + R2 + R3)

Reference: Past Exam Paper – March 2016 Paper 12 Q34

Solution:

Answer: D.

The sum of p.d. across the resistors in the circuit is equal to the overall e.m.f. in the circuit.

The current I is shown to flow from the positive terminal of cell V₁. So, cell V₁ is taken as the reference.

For cells to be connected correctly, the positive terminal of one cell should be connected to the negative terminal of the other. But the negative terminal of V₃ is connected to the negative terminal of V₁ and also, the positive terminal of V₂ is connected to the positive terminal of V₁. Thus, both V₂ and V₃ are wrongly connected to V₁ and hence, the two cells are reducing.

 

Overall p.d. = V₁ – V₂ – V₃ = I(R₁ + R₂ + R₃)