Question

Draw a schematic diagram of a circuit consisting of a battery of $3\mathrm{cells}$ of $2\mathrm{V}$ each, a combination of three resistors of $10\mathrm{\Omega },20\mathrm{\Omega }$, and connected in parallel, a plug key, and an ammeter, all connected in series. Use this circuit to find the value of the current through each resistor.

Answer 1

Schematic diagram of the circuit:

Step 1: Given data

Resistance, $$\begin{gathered} {\mathrm{R}}_1=10\mathrm{\Omega } \\ {\mathrm{R}}_2=20\mathrm{\Omega } \\ {\mathrm{R}}_3=30\mathrm{\Omega } \end{gathered}$$

Number of cells $=3$

The voltage of each cell $=2\mathrm{V}$

The total voltage of the battery, $$\begin{gathered} \mathrm{V}=2\times 3 \\ =6\mathrm{V} \end{gathered}$$

Step 2: Calculating the current through each resistor:

Let, the current is ${\mathrm{I}}_1,{\mathrm{I}}_2,$ and ${\text{I}}_3$ in the resistance ${\mathrm{R}}_1,{\mathrm{R}}_2$_, and ${\mathrm{R}}_3$.

We know that the voltage (V) remains the same across each resistor (R) in a parallel combination.

The formula: $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Substituting the given values we get-

$$\begin{gathered} \Rightarrow {\mathrm{I}}_1=\frac{\mathrm{V}}{{\mathrm{R}}_1} \\ =\frac{6}{10} \\ =0.6\mathrm{A} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{I}}_2=\frac{\mathrm{V}}{{\mathrm{R}}_2} \\ =\frac{6}{20} \\ =0.3\mathrm{A} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{I}}_3=\frac{\mathrm{V}}{{\mathrm{R}}_3} \\ =\frac{6}{30} \\ =0.2\mathrm{A} \end{gathered}$$

Therefore, the current through resistor $R_1,R_2,and{R}_3$ is $0.6\mathrm{A},0.3\mathrm{A},and0.2\mathrm{A}$ respectively.