Question

The electric current through the $10\Omega$ resistor in the circuit given below is ?

Answer 1

Hint: In order to solve this question we need to understand Kirchhoff’s Law for loops. Kirchhoff’s law for junction states that no charges could accumulate at the junction so the net current entering at the junction must leave the junction by the same amount, this law is supported by the equation of continuity at the junction. And the second law states that in a closed loop or in a closed circuit the net potential difference would be zero.

Complete step by step answer:
Before solving let us first solve the loop in upper half containing two emf in same direction
So by applying formula for net emf we get,
$E = \dfrac{{{E_1}{r_1} + {E_2}{r_2}}}{{{r_1} + {r_2}}}$
Using ${E_1} = 10\,V$
${E_2} = 10\,V$
$\Rightarrow {r_1} = 1\Omega $
$\Rightarrow {r_2} = 1\Omega $
We get,
$E = \dfrac{{(10 \times 1) + (10 \times 1)}}{{1 + 1}} \\
\Rightarrow E= 10\,V$

For equivalent resistances in parallel combination using formula ${r_{eq}} = \dfrac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}$ we get,
${r_1} = {r_2} = 1\Omega $
$\Rightarrow {r_{eq}} = \dfrac{{(1 \times 1)}}{{1 + 1}} = 0.5\Omega $
Now using Kirchhoff’s law in closed circuit we get
$E - i{r_{eq}} - iR = 0$
$\Rightarrow i = \dfrac{E}{{{r_{eq}} + R}}$
Using $R = 10\Omega $
And putting values in equation $i = \dfrac{E}{{{r_{eq}} + R}}$ we get
$i = \dfrac{{10}}{{0.5 + 10}} \\
\therefore i = 0.952A$

So the current through $10\Omega $ resistor is $0.952\,A$.

Note: It should be remembered that here the battery is non ideal and that’s why it has non zero internal resistance. If the battery is ideal then the internal resistance of the battery would be zero. Also we use general convention that if we move in a loop in the same direction as current the potential difference across resistance would be taken as negative and if we move in the opposite direction then potential difference across resistance would be taken as positive. Also when we start moving in a loop we first encounter the positive side of the battery so we have taken E as positive otherwise it would be negative.