Suppose current flowing in the circuit is I, then the current passing through resistance R₁ will be I₁, current passing through resistance R₂ will be I₂ and current passing through resistance R₃ will be I₃.

Total current I = \(I_1 + I_2 + I_3\)

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit, we get that

I = \(\frac{V}{R}\)

Since the potential difference across all the resistance is same, so applying the Ohm's law to each resistance we get that

\(I_1 = \frac{V}{R_1}\)

\(I_2 = \frac{V}{R_2}\)

\(I_3 = \frac{V}{R_3}\)

Putting these eqs. in the above one, we get

\( \frac{V}{R}\) = \(\frac{V}{R_1}\) + \(\frac{V}{R_2}\) + \(\frac{V}{R_3}\)

\( \frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

If two resistance are connected in parallel, then the resultant resistance will be

\( \frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

(b) If switch is open, then only upper two resistances (connected in parallel) are in the circuit.

Effective resistance is \( \frac{1}{R_{eq}}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\) = \(\frac{2}{R}\)

\( \frac{1}{R_{eq}}\) = \(\frac{2}{R}\)

so the current = I = \(\frac{V}{\frac{R}{2}}\) = 0.64 (given)

\(\frac{V}{R}\) = 0.3 A

When the switch closes, the third resistance also comes in the circuit. The effective resistance of the circuit becomes \(\frac{R}{3}\)

Hence, Current I = \(\frac{V}{\frac{R}{3}}\) = 3(V/R) = 3 x 0.3 = 0.9 A