Question

In the given circuit, the current through branch having indicated resistor $R_4$ is $375\times {10}^{-x}A$. Find x.

Given $R_1=R_2=4\Omega ,R_3=R_4=R_5=2\Omega ,{\epsilon }_1=5V,{\epsilon }_2=10V$

Answer 1

$R_1=4\Omega$, $R_2=4\Omega$, $R_3=2\Omega$, $R_4=2\Omega$, $R_5=5\Omega$

$E_1=5volt$, $E_2=10volt$

Assign the potential at each node

Let at node $d$ voltage =$OV$

Let at node $c$ voltage =$xV$

Let at nod $e$ voltage = $(x+5)V$

Let at node $B$ voltage =$xV$

Let at node $a$ voltage =$OV$

At node $e$

$I_2=I_3+I_4$ ( Kirchoff's Junction law)

where $I_3=\frac{(x+5)-x}{2}{I}_4=\frac{(x+5)-10}{2}$

$I_2=\frac{5}{2}+\frac{(x-5)}{2}$

Note that outgoing current at node $e$=incoming current at node $b$.

Thus $\frac{(x+5)}{2}+\frac{(0-x)}{4}=I_2+\frac{(x-0)}{4}\to \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$x=-2.5V$

Current through $R_4$

$I_4=\frac{x+5-10}{2}$

$=3.75A$

$=375\times {10}^{-2}A$

Given $375\times {10}^{-x}=375\times {10}^{-2}$

On comparing $x=2$