Question

Three resistances of 4Ω, 8Ω 𝑎𝑛𝑑 16Ω are connected in a parallel to a 220 V supply. Calculate the equivalent resistance and current in circuit?

Answer 1

The equivalent resistance $R$ of a parallel combination of resistors $R_1, R_2,R_3$ is given by

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Given $R_1=4\Omega, R_2=8\Omega, R_3=16\Omega$

$\therefore$ The equivalent resistance of the given circuit is

$\frac{1}{R}=\frac{1}{4\Omega}+\frac{1}{8\Omega}+\frac{1}{16\Omega}\\ \Rightarrow \frac{1}{R}=\frac{4+2+1}{16\Omega}\\ \Rightarrow \frac{1}{R}=\frac{7}{16\Omega}\\ \Rightarrow R=\frac{16}{7}\Omega\\ \Rightarrow R=2.286\Omega$

The current in the circuit is $I=\frac{V}{R}=\frac{220V}{2.286\Omega}=96.24A$

Answer: Equivalent resistance = $2.286\Omega$ , current in the circuit = $96.24A$