Four resistances are connected as shown below.

A potential difference is applied between points P and S.
(a) Calculate the equivalent resistance between points Q and R.
(b) What is the equivalent resistance between points P and S ?
(c) If the current flowing through $3 Ω$ resistance is 2 A, what is the current through $2 Ω$ resistance?
(d) What is the current through the $4 Ω$ resistance?

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Solution

part a: 2 Marks
part b, c, and d: Each 1 Mark

(a) Resistance between Q and R,
$\frac{1}{R_{1}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
Or $R_{1} = 1.2 Ω$
(b) Resistance between P and S,
$R_{2} = 1 + 1.2 + 4 = 6.2 Ω$
(c) Potential difference across $3 Ω$ resistance, $V_{2} = I_{2} R = 2 \times 3 = 6 V$
The above potential difference is also the potential difference across $2 Ω$ resistance as it is connected in parallel with $3 Ω$ resistance between Q and R.
$∴ V_{1} = 2 I_{1} o r 6 = 2 I_{1} o r I_{1} = 3 A$
(d) Total current through $4 Ω$ resistance,
$I = I_{1} + I_{2} = 3 + 2 = 5 A$

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