Question

(a) Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R₁, R₂ and R₃ joined in parallel.
(b) In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistor are identical.

Answer 1

(a)



Let the resistance of the three resistors be R₁, R₂ and R₃, respectively. Let their combined resistance be R. Let the total current flowing in the circuit be I and the strength of the battery be V. Then from Ohm's law, we have:
V = IR .....(1)
We know that when the resistors are connected in parallel, the potential drop across each resistance is the same.
Therefore:
I = I₁ + I₂ + I₃
I = V/R₁ + V/R₂ + V/R₃
I = V/(1/R₁ + 1/R₂ + 1/R₃) ...... (2)
From equations (1) and (2) we have:
1/R = 1/R₁ + 1/R₂ + 1/R₃

(b) Let V be the voltage of the cell and R is the resistance of each resistor. When switch K is not closed then

$$\begin{gathered} \frac{V}{{\displaystyle \frac{R}{2}}}=0.6.....\left(1\right)[\mathrm{Using}V\mathit{}=IR] \\ \mathrm{Let}A\mathrm{be}\mathrm{the}\mathrm{current},\mathrm{when}\mathrm{switch}\left(\mathrm{K}\right)\mathrm{is}\mathrm{closed} \\ \frac{V}{{\displaystyle \frac{R}{3}}}=A.....\left(2\right) \\ \mathrm{Dividing}\mathrm{equation}\left(2\right)\mathrm{by}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{A}{0.6}=1.5 \\ A=1.5\times 0.6=0.9\mathrm{A} \end{gathered}$$