As the switch S is closed in the circuit show | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $V$ be the potential at node $C$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$i_{1} + i_{2} = i_{3}$$\frac{10 - V}{4} + \frac{5 - V}{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $V$ be the potential at node $C$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$i_{1} + i_{2} = i_{3}$$\frac{10 - V}{4} + \frac{5 - V}{2} = \frac{V - 0}{2}$Multiplying the entire equation by $4$ to clear the denominators:$(10 - V) + 2(5 - V) = 2V$$10 - V + 10 - 2V = 2V$$20 - 3V = 2V$$5V = 20$$V = 4 \, V$Now,the current passing through the switch $S$ is $i_{3}$:$i_{3} = \frac{V - 0}{2} = \frac{4 - 0}{2} = 2 \, A$

As the switch $S$ is closed in the circuit shown in the figure,the current passing through it is ............. (in $, A$)

Similar Questions

The figure shows a network of five resistances and two batteries. The current through the $30\,V$ battery is ............... $A$.

In the circuit shown in the figure,find the current through the branch $BD$ in $A$.

Resistances are joined as shown in the figure. In the balanced condition, the current $I$ drawn from the battery is (in $\text{ A}$)

For the given circuit,the reading of the ammeter in amperes is........

In the arrangement shown in the figure,the current through the $5\,\Omega$ resistor is ............. $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module