Question

Three resistors are connected to a $12V$ battery as shown in the figure given below :

1. What is current through $8\mathrm{ohm}$ resistor?
2. What is the potential difference across the parallel combination of $6\mathrm{\Omega }$ and $12\mathrm{\Omega }$ resistor?
3. What is the current through $6\mathrm{\Omega }$ resistor?

Answer 1

Step-1 Given data and drawing diagram

$E.m.f.$of battery $E=12\mathrm{V}$

Let resistance $$\begin{gathered} R_1=8\Omega \\ {R}_2=12\Omega \\ {R}_3=6\Omega \end{gathered}$$

Step-2 Finding equivalent resistances

Resistors of $6\mathrm{\Omega }$and $12\mathrm{\Omega }$are connected in parallel combination.

So, $\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{R_2}+\frac{1}{R_3}$

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{12}+\frac{1}{6} \\ \frac{1}{R_p}=\frac{2+1}{12} \\ {R}_P=4\Omega \end{gathered}$$

Also, resistors $R_{1=}8\mathrm{\Omega }$and ${\mathrm{R}}_{\mathrm{p}}$are connected in a series combination.

So, ${\mathrm{R}}_{\mathrm{eq}}=R_1+R_p$

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}=8+{\mathrm{R}}_{\mathrm{p}} \\ {\mathrm{R}}_{eq}=4+8 \\ {\mathrm{R}}_{eq}=12\mathrm{\Omega } \end{gathered}$$

Step-3 Finding current

Current in the circuit

$\mathrm{I}=\frac{Nete.m.f.}{{\mathrm{R}}_{\mathrm{eq}}}$

$\mathrm{I}=\frac{12}{12}=1\mathrm{A}$

Hence, the current through $8\mathrm{\Omega }$ is $1\mathrm{A}.$

Current through $6\mathrm{\Omega }$ the resistor by using the formula ${\mathrm{I}}_1=\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}\right)\times I$

$$\begin{gathered} {\mathrm{I}}_3=\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_3+{\mathrm{R}}_2}\right)\times I \\ {\mathrm{I}}_3=\frac{12}{12+6}\times 1 \\ {\mathrm{I}}_3=\frac{2}{3}\mathrm{A} \\ {\mathrm{I}}_3=0.66\mathrm{A} \end{gathered}$$

Step-4 Finding potential difference

The potential difference across the parallel combination of $6\mathrm{\Omega }$ and $12\mathrm{\Omega }$ resistor is

$$\begin{gathered} \mathrm{V}=\mathrm{I}{\mathrm{R}}_{\mathrm{p}}(\mathrm{Using}\mathrm{Ohm}\text{'}\mathrm{s}\mathrm{law}) \\ \mathrm{V}=1\times 4 \\ \mathrm{V}=4\mathrm{volts} \end{gathered}$$

Hence, (i) Current through $8\mathrm{\Omega }$ the resistor is $1A$

(ii) Potential difference across the combination of $6\mathrm{\Omega }$ and $12\mathrm{\Omega }$ resistor is $4\mathrm{volts}.$

(iii) Current through $6\mathrm{\Omega }$ a resistor is $0.66A.$