Which one of the switches, S₁, S₂ or S₃, should be opened so that:

(a) total resistance is equal to 4 ohms.

(b) the total resistance is equal 3 ohms.

(c) the current through 4 ohms is 3⁄5^th of the total current.

(a) S3

(b) S1

(c) S2

Explanation:

Let R₁= 6 Ω, R₂= 2 Ω, R₃= 10 Ω, R₄= 4 Ω

If S₃ is open: then R₂ and R₃ are in series

$\text R${\text {s}}= \text R{2} + \text R{3}\\[1em] \text R{\text {s}} = 2\ Ω + 10\ Ω = 12\ Ω

then R_s and R₁ are in parallel

$\dfrac{1}{R''$p}=\dfrac{1}{R}s +\dfrac {1}{R}1 \\[1em] \dfrac{1}{R''p}=\dfrac{1}{12} +\dfrac {1}{6} \\[1em] \dfrac{1}{R''p}=\dfrac{1+2}{12} \\[1em] \dfrac{1}{R''p}=\dfrac{3}{12} \\[1em] \text {R}''p=\dfrac{12}{3} \\[1em] \text {R}''p=4\ Ω \\[1em]

Total resistance is 4 Ω

If S₁ is open: then R₂ and R₃ are in series

$\text R${\text {s}}= \text R{2} + \text R{3}\\[1em] \text R{\text {s}} = 2\ Ω + 10\ Ω = 12\ Ω

and R_s and R₄ are in parallel:

$\dfrac{1}{R$p}=\dfrac{1}{R}s +\dfrac {1}{R}4 \\[1em] \dfrac{1}{Rp}=\dfrac{1}{12} +\dfrac {1}{4} \\[1em] \dfrac{1}{Rp}=\dfrac{3+1}{12} \\[1em] \dfrac{1}{Rp}=\dfrac{4}{12} \\[1em] \text {R}p=\dfrac{12}{4} \\[1em] \text {R}p = 3\ Ω \\[1em]

Total resistance is 3 Ω

If S₂ is open: then R₁ and R₄ are in parallel

$\dfrac{1}{R'$p}=\dfrac{1}{R}1 +\dfrac {1}{R}4 \\[1em] \dfrac{1}{R'p}=\dfrac{1}{6} +\dfrac {1}{4} \\[1em] \dfrac{1}{R'p}=\dfrac{2+3}{12} \\[1em] \dfrac{1}{R'p}=\dfrac{5}{12} \\[1em] \text {R}'p=\dfrac{12}{5} \\[1em] \text {R}'p=2.4\ Ω \\[1em]

Total resistance is 2.4 Ω

Total current drawn in circuit $=\text I =\dfrac{V}{R'}_p = \dfrac{24}{2.4}=10\ \text A$

Voltage across 4 Ω is 24 V

so, current through 4 Ω is $\text I_4 =\dfrac{24}{4}=6\ \text A$

$\text I_4 =\dfrac{3 \text I}{5} =\dfrac{3 \times 10}{5}= 6\ \text A$

So, the current through 4 ohms is 3⁄5^th of the total current.