In the circuit shown, current flowing through $25 V$ cell in amperes is: (answer the nearest integer)

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Solution

Given $E_{1} = 10 V E_{2} = 5 V E_{3} = 20 V E_{4} = - 20 V$
$r_{1} = 5 Ω r_{2} = 10 Ω r_{3} = 5 Ω r_{4} = 11 Ω$
Equivalent resistance: $\frac{1}{r_{e q}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} + \frac{1}{r_{4}}$
$\Rightarrow \frac{1}{r_{e q}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{5} + \frac{1}{11}$
$\Rightarrow r_{e q} = 1.69 Ω$
Equivalent voltage, $E_{e q} = [\frac{E_{1}}{r_{1}} + \frac{(E_{2})}{r_{2}} + \frac{E_{3}}{r_{3}} + \frac{(E_{4})}{r_{4}}] r_{e q}$
$\Rightarrow E_{e q} = [\frac{10}{5} + \frac{(5)}{10} + \frac{20}{5} + \frac{(- 20)}{11}] 1.69$
$\Rightarrow E_{e q} = 7.91 V$
Using Kirchhof's law of second loop, $- 25 + 7.91 + i (1.69) = 0$

$\Rightarrow i = 10.11 A$

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