Question

(a) With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.
(b) In the circuit diagram shown below, find:

(i) Total resistance.
(ii) Current shown by the ammeter A

Answer 1

(a)

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

(b)
(1)
$$\begin{gathered} \mathrm{As}\mathrm{shown}\mathrm{Resistance}2\mathrm{\Omega }\mathrm{and}3\mathrm{\Omega }\mathrm{are}\mathrm{in}\mathrm{series} \\ \mathrm{we}\mathrm{know}\mathrm{resistance}\mathrm{in}\mathrm{series}\mathrm{arrangement}\mathrm{can}\mathrm{be}\mathrm{obtained}\mathrm{as}R=R_1+R_2 \\ R=2\mathrm{\Omega }+3\mathrm{\Omega }=5\mathrm{\Omega } \\ \mathrm{Since}\mathrm{this}5\mathrm{\Omega }\mathrm{is}\mathrm{in}\mathrm{parallel}\mathrm{with}\mathrm{another}5\mathrm{\Omega }\mathrm{resistance}\mathrm{therefore}\mathrm{total}\mathrm{resistance}\mathrm{can}\mathrm{be}\mathrm{obtained}\mathrm{as} \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\mathrm{Here}{\mathrm{R}}_1=5\mathrm{\Omega },{\mathrm{R}}_2=5\mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{5}+\frac{1}{5} \\ \frac{1}{R}=\frac{2}{5} \\ R=2.5\mathrm{\Omega } \\ \mathrm{Total}\mathrm{resistance}\mathrm{of}\mathrm{combination}=2.5\mathrm{\Omega } \end{gathered}$$

(ii) The current shown by the ammeter A, i.e. the current in the circuit can be calculated as:
I = $\frac{\mathrm{V}}{\mathrm{R}}$

I =$\frac{4}{2.5}$= 1.6 A