Question

In the circuit shown in figure, $E_1$ and $E_2$ are two ideal sources of unknown emfs. Current flowing through resistor of resistance $R\Omega$ is $2A$ and through resistor of resistance $4\Omega$ is $3A$ as shown. If potential difference across $6\Omega$ resistance is $V_A-V_B=10\text{V}$, then the value of $(E_2-E_1)R$( in $V.\Omega$) is

Answer 1

Since, $V_A-V_B=10\text{V}\Rightarrow i_1\left(6\right)=10\Rightarrow i_1=\frac{5}{3}A$

As AB is parallel to ED, potential difference across ED is also $10V$. So,
$i_2\left(3\right)=10\text{V}\Rightarrow i_2=\frac{10}{3}\text{A}\Rightarrow i_1+i_2=5A$ and
From Kirchhoff's junction law current through CF is
$3+i_1+i_2=8A$

Now applying KVL to the loop EFCDE
$-E_2+8\left(3\right)+4\left(5\right)+3\left(\frac{10}{3}\right)=0$
$\Rightarrow E_2=54\text{V}$
Applying KVL to the loop FGHCF
$E_1-4\left(3\right)-8\left(3\right)=0$
$\Rightarrow E_1=36\text{V}$
Applying KVL to the loop AEFGIJA
$-E_2+E_1+2R=0$
$2R=54-36$
$\Rightarrow R=9\Omega$
Numerical value of
$(E_2-E_1)R=162V\Omega$