In the following circuit,the current I_3 is | vedclass

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

Question

(D) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the central node:$I_1 + I_2 = I_3$Using Ohm's law for ea

Vedclass · Accepted Answer

$-\frac{5}{6} \ A$

Vedclass · Answer

(D) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the central node:$I_1 + I_2 = I_3$Using Ohm's law for each branch:$I_1 = \frac{8 - 6 - V}{28} = \frac{2 - V}{28}$$I_2 = \frac{12 - 6 - V}{54} = \frac{6 - V}{54}$$I_3 = \frac{V - 0}{0} = 	ext{undefined (this approach is complex)}$.Alternatively,using nodal analysis at the central node:$\frac{V - 8}{28} + \frac{V - 12}{54} + \frac{V - 6}{0} = 0$ is not correct. Let's use loop analysis.For loop $1$: $28 I_1 + 6 + 8 = 0 \implies 28 I_1 = -14 \implies I_1 = -0.5 \ A$.For loop $2$: $54 I_2 + 6 + 12 = 0 \implies 54 I_2 = -18 \implies I_2 = -1/3 \ A$.By $KCL$ at the junction: $I_3 = I_1 + I_2 = -0.5 + (-1/3) = -1/2 - 1/3 = -5/6 \ A$.

In the following circuit,the current $I_3$ is

Similar Questions

In the given circuit,the value of $i$ is ........... $A$.

In the circuit shown in the figure,the current drawn from the battery is $4 \, A$. If the $10 \, \Omega$ resistor is replaced by a $20 \, \Omega$ resistor,then the current drawn from the circuit will be .............. $A$.

In the circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $Omega$)

Five resistors are connected as shown in the diagram. The equivalent resistance between $A$ and $B$ is .............. $\Omega$.

In the network shown,the cell of emf $E$ has internal resistance $r$ and the galvanometer shows zero deflection. If the cell is replaced by a new cell of emf $2E$ and internal resistance $3r$,keeping everything else identical,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module