91Ó°ÊÓ

i) The resistance $1$ is $R_1=100\text{‰ө}$

ii) The resistance $2$ and resistance $3$ are $R_2=R_3=50.0\text{‰ө}$

iii) The resistance $4$ is $R_4=75.0\text{‰ө}$

iv) The emf of the battery is $\mathrm{Î\mu }=6.00\text{ V}$

We can use the concept ofjunction law and loop law. The algebraic sum of potential across any electric component and electric potential is zero. The sum of the currents entering the junction must be equal to the sum of the currents leaving the junction. We can use Ohm’s law.

Formula:

$\mathit{R}\mathbf{=}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{4}}}{{\mathbf{R}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{3}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{4}}\mathbf{+}{\mathbf{R}}_{\mathbf{3}}{\mathbf{R}}_{\mathbf{4}}}$

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{R}}_{\mathbf{1}}\mathbf{+}\mathit{R}$

$\mathit{V}\mathbf{=}\mathit{I}\mathit{R}$

The equivalent resistance:

According to the circuit diagram, the resistances$R_2$,$R_{3,}$and$R_4$are in parallel, and$R$its equivalent resistance. Then

$\begin{array}{c}\frac{1}{R}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\ R=\frac{R_2{R}_3{R}_4}{R_2{R}_3+R_2{R}_4+R_3{R}_4}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}R=\frac{50\text{‰ө}×50.0\text{‰ө}×75.0\text{‰ө}}{50.0\text{‰ө}×50.0\text{‰ө}+50.0\text{‰ө}×75.0\text{‰ө}+50.0\text{‰ө}×75.0\text{‰ө}}\\ R=18.75\text{‰ө}\end{array}$

The resistance$R$and$R_1$are in series, then the equivalent resistance of the circuit will be

$R_{eq}=R_1+R$

Substitute all the value in the above equation.

$\begin{array}{c}{R}_{eq}=100\text{‰ө}+18.75\text{‰ө}\\ R_{eq}=118.75\text{‰ө}\\ R_{eq}≃119\text{‰ө}\end{array}$

Hence the equivalent resistance is $R_{eq}=119\text{‰ө}$

The current in resistance 1 :

According to Ohm’s law,

$i_1=\frac{\mathrm{Î\mu }}{R_{eq}}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_1=\frac{6.00\text{ V}}{118.8\text{‰ө}}\\ i_1=5.05×{10}^{−2}\text{‼î}\end{array}$

Hence the current in resistance 1 is $i_1=5.05×{10}^{−2}\text{‼î}$

The current in resistance 2:

Consider loop $ABCD$as

According to the loop law,

$\begin{array}{c}−i_1{R}_1−i_2{R}_2+\mathrm{Î\mu }=0\\ i_2=\frac{\mathrm{Î\mu }−i_1{R}_1}{R_2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_3=\frac{6.00\text{ V}−5.05×{10}^{−2}\text{‼î}×100\text{‰ө}}{50.0\text{‰ө}}\\ i_3=1.90×{10}^{−2}\text{‼î}\end{array}$

Hence the current in resistance 3 is $i_2=1.90×{10}^{−2}\text{‼î}$

The current in resistance:

Consider loop$ABCD$ as

According to the loop law,

$\begin{array}{c}−i_1{R}_1−i_3{R}_3+\mathrm{Î\mu }=0\\ i_3=\frac{\mathrm{Î\mu }−i_1{R}_1}{R_3}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_3=\frac{6.00\text{ V}−5.05×{10}^{−2}\text{‼î}×100\text{‰ө}}{50.0\text{‰ө}}\\ i_3=1.90×{10}^{−2}\text{‼î}\end{array}$

Hence the current in resistance 3 is $i_2=1.90×{10}^{−2}\text{‼î}$

The current in resistance:

According to the junction law,

$\begin{array}{c}−i_1+i_2+i_3+i_4=0\\ i_4=i_1−i_2−i_3\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_4=5.05×{10}^{−2}\text{‼î}−1.90×{10}^{−2}\text{‼î}−1.90×{10}^{−2}\text{‼î}\\ i_4=1.25×{10}^{−2}\text{‼î}\end{array}$

Hence the current in resistance $4$ is $i_1=1.25×{10}^{−2}\text{‼î}$