Question

In the circuit shown, current flowing through $25V$ cell is:

• A. $7.2A$
• B. $10A$
• C. $12A$
• D. $14.2A$

Answer 1

Answer: (C)

$12A$

Given : $E_1=10V$ $E_2=5V$ $E_3=20V$ $E_4=30V$

$r_1=5\Omega$ $r_2=10\Omega$ $r_3=5\Omega$ $r_4=11\Omega$

Equivalent resistance: $\frac{1}{r_{eq}}=$ $\frac{1}{r_1}+$ $\frac{1}{r_2}+$ $\frac{1}{r_3}+$ $\frac{1}{r_4}$

$\Rightarrow \frac{1}{r_{eq}}=$ $\frac{1}{5}+$ $\frac{1}{10}+$ $\frac{1}{5}+$ $\frac{1}{11}$

$\Rightarrow r_{eq}=1.69\Omega$

Equivalent voltage, $E_{eq}=[$ $\frac{E_1}{r_1}+$ $\frac{(-E_2)}{r_2}+$ $\frac{E_3}{r_3}+$ $\frac{(-E_4)}{r_4}]r_{eq}$

$\Rightarrow E_{eq}=[$ $\frac{10}{5}+$ $\frac{(-5)}{10}+$ $\frac{20}{5}+$ $\frac{(-30)}{11}]1.69$

$\Rightarrow E_{eq}=4.68$ $V$

Using Kirchhof's law for second loop, $-25+4.68+i\left(1.69\right)=0$

$⟹$ $i=12$ $A$