Find the current flowing through the resistan | vedclass

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

Question

(D) Let $V_0$ be the potential at point $O$. Applying the nodal analysis method at node $O$:$\frac{V_0 - V_1}{R_1} + \frac{V_0 - V_2}{R_2} + \frac{V_0

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $V_0$ be the potential at point $O$. Applying the nodal analysis method at node $O$:$\frac{V_0 - V_1}{R_1} + \frac{V_0 - V_2}{R_2} + \frac{V_0 - V_3}{R_3} = 0$Substituting the given values:$\frac{V_0 - 20}{20} + \frac{V_0 - 30}{30} + \frac{V_0 - 60}{60} = 0$Multiplying the entire equation by $60$ to clear the denominators:$3(V_0 - 20) + 2(V_0 - 30) + 1(V_0 - 60) = 0$$3V_0 - 60 + 2V_0 - 60 + V_0 - 60 = 0$$6V_0 - 180 = 0$$6V_0 = 180 \implies V_0 = 30 \ V$The current $i_2$ flowing through resistance $R_2$ is given by:$i_2 = \frac{V_0 - V_2}{R_2} = \frac{30 - 30}{30} = 0 \ A$

Find the current flowing through the resistance $R_2$ of the circuit shown in the figure,if the resistances are $R_1 = 20 \ \Omega, R_2 = 30 \ \Omega$ and $R_3 = 60 \ \Omega$,and the potentials of points $1, 2$ and $3$ are $V_1 = 20 \ V, V_2 = 30 \ V$ and $V_3 = 60 \ V$ respectively. (in $A$)

Similar Questions

Which of the following is not a correct statement?

In the given circuit,the potentials at points $B$,$C$,and $D$ will be:

What will happen when a $40\, W$,$220\, V$ lamp and a $100\, W$,$220\, V$ lamp are connected in series across a $40\, V$ supply?

Two cities are $150\, km$ apart. Electric power is sent from one city to another city through copper wires. The fall of potential per $km$ is $8\, V$ and the average resistance per $km$ is $0.5\, \Omega$. The power loss in the wire is

In the circuit shown below,the cell has an $e.m.f.$ of $10\,V$ and internal resistance of $1\,\Omega$. The other resistances are shown in the figure. The potential difference ${V_A} - {V_B}$ is ................ $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module