The figure below shows a portion of an electr | vedclass

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

Question

(D) To find the current in the portion $PQ$,we apply Kirchhoff's current law $(KCL)$ at the junctions.$1$. At the junction where the $2 \, A$ and $8 \

Vedclass · Accepted Answer

$6 \, A$ from $Q$ to $P$

Vedclass · Answer

(D) To find the current in the portion $PQ$,we apply Kirchhoff's current law $(KCL)$ at the junctions.$1$. At the junction where the $2 \, A$ and $8 \, A$ currents meet,the total incoming current is $2 \, A + 8 \, A = 10 \, A$. This $10 \, A$ current flows towards the next junction.$2$. At the junction below $Q$,the total outgoing current is $4 \, A + 2 \, A = 6 \, A$. Since $10 \, A$ enters this junction and $6 \, A$ leaves downwards,the remaining $10 \, A - 6 \, A = 4 \, A$ must flow upwards towards $Q$.$3$. At junction $Q$,the incoming currents are $3 \, A$ (from the top branch) and $4 \, A$ (from the lower branch). The outgoing current is $1 \, A$ (to the right). Let the current in $PQ$ be $I_{PQ}$ flowing from $Q$ to $P$.$4$. Applying $KCL$ at junction $Q$: $I_{incoming} = I_{outgoing}$.$3 \, A + 4 \, A = 1 \, A + I_{PQ}$$7 \, A = 1 \, A + I_{PQ}$$I_{PQ} = 6 \, A$.Since the result is positive,the current $6 \, A$ flows in the assumed direction,i.e.,from $Q$ to $P$.

The figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion $PQ$ is

Similar Questions

Kirchhoff's first law,$i.e.$,$\Sigma i = 0$ at a junction,is based on the law of conservation of:

In the given circuit,if no current flows through the ammeter,what is the value of $X$ in $\Omega$?

In the following circuit,the current $I_3$ is

In the given circuit, what is the potential difference between $X$ and $Y$ in $V$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module