The figure shows a part of an electric circui | vedclass

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

Question

(A) Let the potential at the junction point be $x$.Applying Kirchhoff's Current Law $(KCL)$ at the junction point,the sum of currents leaving the junc

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) Let the potential at the junction point be $x$.Applying Kirchhoff's Current Law $(KCL)$ at the junction point,the sum of currents leaving the junction must be zero:$\frac{x-30}{10} + \frac{x-12}{20} + \frac{x-2}{30} = 0$Multiplying the entire equation by $60$ to clear the denominators:$6(x-30) + 3(x-12) + 2(x-2) = 0$$6x - 180 + 3x - 36 + 2x - 4 = 0$$11x - 220 = 0$$11x = 220$$x = 20\,V$Now,the current through the $20\,\Omega$ resistor is given by:$I = \frac{x - 12}{20}$$I = \frac{20 - 12}{20} = \frac{8}{20} = 0.4\,A$.

The figure shows a part of an electric circuit. The potentials at points $a$,$b$,and $c$ are $30\,V$,$12\,V$,and $2\,V$ respectively. The current through the $20\,\Omega$ resistor will be $........\,A$.

Similar Questions

If in the experiment of Wheatstone's bridge,the positions of cells and galvanometer are interchanged,then the balance point will

In the given circuit,the values of $I_1, I_2, I_3$ are respectively:

Write the necessary facts to understand Kirchhoff's laws.

In a balanced Wheatstone's network,the resistances in the arms $Q$ and $S$ are interchanged. As a result of this:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module