In the cube shown,current i enters at H and l | vedclass

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

Question

(B) Applying Kirchhoff's junction law at node $H$:$i_{HG} = i - (i_{HA} + i_{HE}) = i - (\frac{i}{2} + \frac{i}{6}) = i - \frac{4i}{6} = \frac{i}{3}$.

Vedclass · Accepted Answer

$FC$

Vedclass · Answer

(B) Applying Kirchhoff's junction law at node $H$:$i_{HG} = i - (i_{HA} + i_{HE}) = i - (\frac{i}{2} + \frac{i}{6}) = i - \frac{4i}{6} = \frac{i}{3}$.Applying junction law at node $G$:$i_{HG} = i_{GF} + i_{GB} \Rightarrow \frac{i}{3} = \frac{i}{6} + i_{GB} \Rightarrow i_{GB} = \frac{i}{6}$.Applying junction law at node $B$:$i_{BC} = i_{AB} + i_{GB} = \frac{i}{6} + \frac{i}{6} = \frac{2i}{6} = \frac{i}{3}$.Applying junction law at node $C$:$i_{BC} + i_{DC} + i_{FC} = i \Rightarrow \frac{i}{3} + \frac{2i}{3} + i_{FC} = i \Rightarrow i + i_{FC} = i \Rightarrow i_{FC} = 0$.Thus,the current in branch $FC$ is zero.

In the cube shown,current $i$ enters at $H$ and leaves at $C$. If $i_{AB} = \frac{i}{6}$,$i_{DC} = \frac{2i}{3}$,$i_{HA} = \frac{i}{2}$,$i_{GF} = \frac{i}{6}$,and $i_{HE} = \frac{i}{6}$,choose the branch in which the current is zero.

Similar Questions

In a balanced Wheatstone bridge,the current in the galvanometer is zero. It remains zero when:

The equivalent resistance between points $A$ and $B$ is ............ $\Omega$.

In the given circuit,find the potential difference between $P$ and $Q$ in $V$.

The potential difference $(V_{A}-V_{B})$ between the points $A$ and $B$ in the given part of the circuit is

When the resistance $R$ (indicated in the figure below) is changed from $1 \, k\Omega$ to $10 \, k\Omega$,the current flowing through the resistance $R'$ does not change. What is the value of the resistor $R'$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module