91Ó°ÊÓ

The value of resistances given in figure 27-54,${\text{R}}_{\text{1}}=1.0{\text{‰ө,R}}_{\text{2}}=2.0\text{‰ө}$

The value of battery emf given in figure 27-54,

For resistors in series connection, the current flowing through them is the same and is given by the net potential difference and the total resistance along the arm. Similarly, resistors in parallel joints, have the same potential across them while different current values in the ratio of the resistance values. Using Kirchhoff's voltage law equation, we can get the net potential difference or any unknown parameter across the battery. Kirchhoff’s voltage law states that in any closed loop network, the total voltage around the loop is equal to zero.

Formulae:

The voltage equation using Ohm’s law,$\text{V}=\text{IR}$(1)

Kirchhoff’s voltage law, $\underset{closedloop}{∑}V=0$ (2)

We note that th${\text{R}}_{\text{1}}$eresistors occur in series pairs, contributing net resistance$2{\text{R}}_{\text{1}}$in each branch where they appear. Since,${\mathrm{Î\mu }}_2={\mathrm{Î\mu }}_3$and${\text{R}}_{\text{2}}=2{\text{R}}_{\text{1}}$, from symmetry, we know that the currents throughandare the same, and thus is given as:$I_2(=I_3)=i$

Now, the current through${\mathrm{Î\mu }}_1$is given by:$I_1=2i$

Now, using equation (1) in equation (2), the net potential difference across the arm ab can be given for both the loops as:

For the first loop, $V_b−V_a={\mathrm{Î\mu }}_2−i{R}_2\mathrm{........................}\left(3\right)$

For the second loop, $V_b−V_a={\mathrm{Î\mu }}_1+\left(2R_1\right)\left(2i\right)\mathrm{...........................}\left(4\right)$

Now, using equations (3) and (4), the current value can be given as:

$\begin{array}{l}{\mathrm{Î\mu }}_2−i{R}_2={\mathrm{Î\mu }}_1+\left(2R_1\right)\left(2i\right)\\ i(4R_1+R_2)={\mathrm{Î\mu }}_2−{\mathrm{Î\mu }}_1\\ i=\frac{{\mathrm{Î\mu }}_2−{\mathrm{Î\mu }}_1}{(4R_1+R_2)}\\ i=\frac{4.0\text{ V}−2.0\text{ V}}{4\left(1.0\text{‰ө}\right)+2.0\text{‰ө}}\\ i=0.33\text{‼î}\end{array}$

Thus, the current through the first battery is given as:

$\begin{array}{c}{I}_1=2\left(0.33\text{‼î}\right)\\ =0.66\text{‼î}\end{array}$

Hence, the current value is .$0.66\text{​‼î}$

The direction of ${\text{I}}_{\text{1}}$is downward.

From calculations of the part (3), the current across battery 2 is given by,

$\begin{array}{c}{I}_2=i\\ =0.33\text{‼î}\end{array}$

Hence, the current value is$0.33\text{‼î}$

The direction of ${\text{I}}_{\text{2}}$is upward.

From calculations of the part (3), the current across battery 3 is given by,

$\begin{array}{c}{I}_3=i\\ =0.33\text{‼î}\end{array}$

Hence, the current value is .$0.33\text{‼î}$

The direction of${\text{I}}_{\text{3}}$ is upward.

From calculations of the part (3), the potential difference across arm ab can be given using equation (3) as follows:

$\begin{array}{c}{V}_b−V_a=4.0\text{ }V−2\left(0.66\text{ V}\right)\\ =2.68\text{ V}\end{array}$

Hence, the potential difference is .$2.68\text{ V}$