91Ó°ÊÓ

1. The resistance value of all the resistors,$\mathrm{R}=4.0\mathrm{Î\copyright }$
2. Emf of the all the ideal batteries,$\mathrm{Î\mu }=4.0\mathrm{V}$

We know that the current flows through the positive to the negative terminal of the battery. In the given problem, we need to create a loop such that there is no resistance line through the loop to cause any external disturbance to the required current value. Thus, we choose such a manner that we end up with only resistor R attached to the ideal batteries that contribute to the net current value through this resistor considering no other resistors.

Formula:

The voltage equation using Ohm’s law,

$\mathbf{V}\mathbf{=}\mathbf{IR}$ …(¾±)

Kirchhoff’s voltage law,

$\underset{\mathbf{closed}\mathbf{}\mathbf{loop}}{\mathbf{∑}}\mathbf{V}\mathbf{=}\mathbf{0}$ …(¾±¾±)

Using equation (ii), we need to choose a loop that only has resistance R to get the required current value. (Since, if we encounter any other resistance, the required current will change as resistor will not be in series with the given resistance)

Thus, we get the voltage equation using equation (i) in equation (ii) as follows for the above circuit as follows:

$\begin{array}{rcl}+\mathrm{Î\mu }+\mathrm{Î\mu }+\mathrm{Î\mu }-\mathrm{Î\mu }-\mathrm{IR}& =& 0\\ \mathrm{I}& =& \frac{2\mathrm{Î\mu }}{\mathrm{R}}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{I}& =& \frac{2\left(4.0\mathrm{V}\right)}{4.0\mathrm{Î\copyright }}\\ & =& 2.0\mathrm{A}\end{array}$

Hence, the current value is, 2.0 A.