The value of current through the 20,Omega res | vedclass

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

Question

(A) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at this node:$\frac{V-10}{10} + \frac{V-(-6)}{20} + \frac{V

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(A) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at this node:$\frac{V-10}{10} + \frac{V-(-6)}{20} + \frac{V}{10} + \frac{V-10}{5} = 0$Multiplying the entire equation by $20$:$2(V-10) + (V+6) + 2V + 4(V-10) = 0$$2V - 20 + V + 6 + 2V + 4V - 40 = 0$$9V - 54 = 0$$V = 6 \,V$Now,the current through the $20\,\Omega$ resistor is given by:$I = \frac{V - (-6)}{20} = \frac{6 + 6}{20} = \frac{12}{20} = 0.6 \,A$.

The value of current through the $20\,\Omega$ resistor is ............ $A$.

Similar Questions

If the charge on the capacitor is $1 \text{ mC}$ in the given circuit,then $\frac{R_1 R_2}{R_3} = \ldots \ldots \ldots \Omega$.

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

Four resistors are connected as shown in the figure. $A$ $6\,V$ battery of negligible resistance is connected across terminals $A$ and $C$. The potential difference across terminals $B$ and $D$ will be ............. $V$.

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

Assertion : Kirchhoff's junction rule follows from the conservation of charge.
Reason : Kirchhoff's loop rule follows from the conservation of momentum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module