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

(A) Let the potential at the junction be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the junction:$\frac{V - 20}{2} + \frac{V - 5}{4} + \frac{V -

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) Let the potential at the junction be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the junction:$\frac{V - 20}{2} + \frac{V - 5}{4} + \frac{V - 0}{2} = 0$Multiplying the entire equation by $4$ to clear the denominators:$2(V - 20) + (V - 5) + 2V = 0$$2V - 40 + V - 5 + 2V = 0$$5V - 45 = 0$$5V = 45$$V = 9 \, V$Now,the current $I_3$ passing through the switch $S$ (which is in series with the $2 \, \Omega$ resistor connected to ground) is:$I_3 = \frac{V - 0}{2} = \frac{9}{2} = 4.5 \, A$.

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

Similar Questions

How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?

The materials having negative magnetic susceptibility are

If the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again at $Q$,then the coordinates of $Q$ are

The points $A(a, 0)$,$B(0, b)$,$C(c, 0)$,and $D(0, d)$ are such that $ac = bd$ and $a, b, c, d$ are all non-zero. Then the points:

In a cyclic process,the work done by the system is

Vedclass Test Series

Exam Paper Generator

Online Exam Module