First slide

Sources of emf and Kirchhoff's laws

Question

In the given circuit [choose the incorrect option]

Easy

A

The current through the $10 Ω$ resistor is 2A
B

The current through the $5 Ω$ resistor is 2.5 A
C

The current through the $25 V$ battery is 6.25 A
D

The p.d between the terminals of 10 V battery is 12.5 V

Solution

$E = \frac{\frac{10}{1} + \frac{25}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{45}{3} = 15 V$
$r = \frac{1 \times 2}{1 + 2} = \frac{2}{3} Ω$

$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{5 (10)}{5 + 10} = \frac{10}{3} Ω i = \frac{15}{\frac{2}{3} + \frac{10}{3}} = \frac{15}{4} A$
$∴ V_{A} {- V}_{B} = i \times \frac{10}{3} = \frac{15}{4} \times \frac{10}{3} = 12.5 V$

Current in $10 Ω$ = $\frac{V_{A} {- V}_{B}}{10} = \frac{12.5}{10} = 1.25 A$
Current in $5 Ω$ = $\frac{V_{A} {- V}_{B}}{5} = \frac{12.5}{5} = 2.5 A$
Current in 25V battery , $V_{A} {- V}_{B} = 25 - 2 i_{0}$
$\Rightarrow 12.5 = 25 - 2 i_{0} \Rightarrow i_{0} = 6.25 A$

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