Question

For the circuit given below, if the current in the $1\Omega$ resistor is $\frac{x}{23}A$ , what is the value of $x$ ? Assume the batteries are ideal.

Answer 1

Answer: 57

The circuit given in the problem can be redrawn as shown below.

The three branches are connected between the same points $P$ and $Q$ and hence they are in parallel. For such a circuit the potential difference between the points $P$ and $Q$ is
$V_{QP}=\frac{\frac{E_1}{R_1}+\frac{E_2}{R_2}+\frac{E_3}{R_3}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$
Where $E_1,E_2\text{and}{E}_3$ are the emf of the batteries present in the first, second and the third branch.
Similarly, $R_1,R_2\text{and}{R}_3$ are the values of resistance in each of the respective branches.
$V_{QP}=\frac{\frac{6}{3}+\frac{0}{1}+\frac{9}{5}}{\frac{1}{3}+\frac{1}{1}+\frac{1}{5}}=\frac{57}{23}V$
The value of current through the $1\Omega$ resistance is
$i=\frac{V_{QP}}{1}=\frac{57}{23}A=2.48A$