A network of resistors, each of resistance R, is shown in Fig. 7.1.

Question 22

(a) A network of resistors, each of resistance R, is shown in Fig. 7.1.

Fig. 7.1

Switches S1 and S2 may be ‘open’ or ‘closed’.

Complete Fig. 7.2 by calculating the resistance, in terms of R, between points X and Y

for the switches in the positions shown.

Fig. 7.2

[3]

(b) Two cells of e.m.f. E1 and E2 and negligible internal resistance are connected into a

network of resistors, as shown in Fig. 7.3.

Fig. 7.3

The currents in the network are as indicated in Fig. 7.3.

Use Kirchhoff’s laws to state the relation

(i) between currents I1, I2, I3 and I4, [1]

(ii) between E1, E2, R, and I3 in loop NKLMN, [1]

(iii) between E2, R, I3 and I4 in loop NKQN. [1]

Reference: Past Exam Paper – June 2009 Paper 22 Q7

Solution:

(a)

∞                                

{When the switches are open, no current would flow. This means that the resistance is infinite.}

(R + R =) 2R              

{Since switch S1 is open, no current flows in the left branch of resistors. These 2 resistors can be neglected in the effective resistance.}

([1/2R + 1/2R]^-1 =) R 

{When both switches are closed, current flows through all the resistors.}

(b)

(i) I₁ + I₃ = I₂ + I₄
{Sum of current entering junction N = Sum of current leaving junction N}

(ii) E₂ – E₁ = I₃R        

{From Kirchhoff’s law, the overall e.m.f. in a loop is equal to the sum of p.d. in the loop.

Considering the loop NKLMN, we have to consider 2 different currents: I₁ and I₃ which are in opposite direction.

I₁ comes from cell E₁ and I₃ from cell E₂. We need to identify which of the two cells is greater in e.m.f.

Since the current flowing through the resistor is I₃ (and not I₁), we can conclude that E₂ is greater than E₁.

This is why we consider the E₂ – E₁ and not E₁ – E₂.}

(iii) E₂ = I₃R + 2I₄R

{From Kirchhoff’s law, the overall e.m.f. in a loop is equal to the sum of p.d. in the loop.}