91Ó°ÊÓ

$\mathrm{Î\mu }=12.0\text{ }V$

$R_1=2000\text{‰ө}$

$R_2=3000\text{‰ө}$

$R_3=4000\text{‰ө}$

We use the loop rule (Kirchhoff’s Voltage law) and the junction rule (Kirchhoff’s Current law) to find the current flowing through each resistor. Then, we can find the potential difference for any combination using Ohm’s law.

Formula:

i) For any loop,$\mathbf{∑}\mathit{V}\mathbf{=}\mathbf{0}$

i) At any junction,$∑I_i=∑I_o$

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

We assume that,

Current flowing through$R_1$is$I_1$ .

Current flowing through$R_2$ is$I_2$ .

Current flowing through$R_3$is.$I_3$

By applying the junction rule, we get

$I_3=I_1−I_2$

First, we consider the loop ABDA,

Using the loop rule, we can write as

$\mathrm{Î\mu }−I_2{R}_2−I_1{R}_1=0$

Substitute all the value in the above equation.

$12\text{ V}−\left(3000\text{‰ө}\right)I_2−\left(2000\text{‰ө}\right)I_1=0$

$\left(2000\text{‰ө}\right)I_1+\left(3000\text{‰ө}\right)I_2=12\text{ V}$ …(1)

Now, we consider the loop ABCDA,

Using the loop rule, we can write

$\begin{array}{c}\mathrm{Î\mu }−I_1{R}_1−I_3{R}_3−I_1{R}_1=0\\ \mathrm{Î\mu }−2I_1{R}_1−(I_1−I_2)R_3=0\\ \mathrm{Î\mu }−(2R_1+R_3)I_1+I_2{R}_3=0\end{array}$

Substitute all the value in the above equation.

$12\text{ V}−(4000\text{‰ө}+4000\text{‰ө})I_1+\left(4000\text{‰ө}\right)I_2=0$ …(2)

$\left(8000\text{‰ө}\right)I_1−\left(4000\text{‰ө}\right)I_2=12\text{ V}$

On solving equationsand, we can get the values of$I_1$and$I_2$

$I_1=2.625×{10}^{−3}\text{‼î}$

$I_2=2.25×{10}^{−3}\text{‼î}$

Therefore,

$I_3=I_1−I_2$

Substitute all the value in the above equation.

$\begin{array}{c}{I}_3=(2.625×{10}^{−3}\text{‼î})−(2.25×{10}^{−3}\text{‼î})\\ =3.75×{10}^{−4}\text{‼î}\end{array}$

Now using Ohm’s law, we can find all the potential differences.

$(V_A−V_B)=I_1{R}_1$

Substitute all the value in the above equation.

$\begin{array}{c}(V_A−V_B)=2.625×{10}^{−3}\text{‼î}×2000\text{‰ө}\\ =5.25\text{ V}\end{array}$

Hence the potential difference is,$5.25\text{ V}$ .

$(V_B−V_C)=I_3{R}_3$

Substitute all the value in the above equation.

$\begin{array}{c}(V_B−V_C)=3.75×{10}^{−4}\text{‼î}×4000\text{‰ө}\\ =1.5\text{ V}\end{array}$

Hence the potential difference is,$1.5\text{ V}$.

$(V_C−V_D)=I_1{R}_1$

Substitute all the value in the above equation.

$\begin{array}{c}(V_C−V_D)=2.625×{10}^{−4}\text{‼î}×2000\text{‰ө}\\ =2.25\text{ V}\end{array}$

Hence the potential difference is,$2.25\text{ V}$ .

$(V_A−V_C)=I_2{R}_2$

Substitute all the value in the above equation.

$\begin{array}{c}(V_C−V_D)=2.25×{10}^{−4}\text{‼î}×3000\text{‰ө}\\ =6.75\text{ V}\end{array}$

Hence the potential difference is,$6.75\text{ V}$ .