A network of resistors, each of resistance 1 Ω, is connected as shown. The current passing through the end resistor is 1 A.

Question 3

A network of resistors, each of resistance 1 Ω, is connected as shown.

The current passing through the end resistor is 1 A.

What is the potential difference (p.d.) V across the input terminals?

A 2 V               B 5 V               C 8 V               D 13 V

Reference: Past Exam Paper – November 2015 Paper 13 Q37

Solution:

Answer: D.

We know that

- When simplifying a circuit into a series connections of resistors, the e.m.f. in the circuit is equal to the sum of p.d. across the components.

- For components connected in parallel to each other, the p.d. across them is equal.

- At any junction, the sum of current entering the junction is equal to current leaving the junction.

To understand the flow of current in the circuit above, the junctions have been marked as A, B, C and D, and the currents have been indicated. The way the circuit above is drawn is a bit tricky and confusing for some. It is easier to draw all the resistors in a horizontal way so that it becomes easier to follow the flow of current.

I will now explain how I arrived at this circuit, which is an equivalent to the circuit given in the question.

Consider junction A. To its left is resistor R₁ and at the junction, the current splits to go into resistor R₂ and R₃. This is why we have 2 branches at A.

R₂, which is the lower branch of the two, is connected to junction C. R₃ is connected between A and B.

At junction B, the current splits into 2 branches; one with a single resistor R₄ and another one with the 2 resistors R₅ and R₆ connected in series. The 2 branches join at junction D.

There is no resistor between C and D. So, the circuit drawn is correct.

Consider the current I₅. It passes through the 2 resistors on the right on BD since the 2 resistors are connected in series. The total resistance in that part is 2 Ω.

p.d. in that branch = IV = 1 × (1+1) = 2V

Now, consider junction B: I₃ = I₄ + I₅

We know that the greater the resistance, the smaller the current flowing. When the total resistance is 2 Ω (as above), the current is 1A. Current I4 passes through only a 1 Ω resistor, so the current would be twice that above. So, I₄ = 2A.

p.d. across R₄ = IV = 2 × 1 = 2 V

This is expected since it is in parallel with the above 2 resistors.

Also, I₃ = I₄ + I₅ = 2 + 1 = 3 A

p.d. across R₃ = IV = 3 × 1 = 3 V

The sum of p.d. across the upper branch AC is 3V + 2V = 5V

For a parallel connection, the sum of p.d. across the junctions are equal. So, the p.d. across R₂ is also 5 V.

Current I₂ flowing through R₂ = V / R = 5 / 1 = 5 A

At junction A: I₁ = I₂ + I₃ = 5 + 3 = 8 A

p.d. across R₁ = IR = 8 × 1 = 8 V

e.m.f. in circuit = sum of p.d. across any loop

e.m.f. in circuit = 8 + 3 + 2 = 13 V                 OR = 8 + 5 = 13 V