Question

Four wires meet at a junction. In two of the wires, currents I1=1.47A and I2=2.75A enter the junction. In on of the wires current I3=6.35A leaves the junction. Find the current in the fourth wire and idicate its direction.
Current: (amps)
Direction: (out of junction, into the junction, undetermined)

Answer 1

Concepts and reason

The concept required to solve the given problem is Kirchhoff’s junction rule.

Initially, determine the magnitude of the current in the fourth wire by using Kirchhoff’s junction rule and taking currents towards junction as positive and currents away from the junction as negative. Finally, determine the direction of current by using obtained sign for current in the fourth wire.

Fundamentals

Kirchhoff’s loop rule states that sum of voltage drop across a closed loop is equal to zero.

$\sum V = 0$

Here, $\sum V$ is the sum of voltage drop.

According to Kirchhoff’s junction rule, the sum of currents meet at a junction is equal to zero.

$\sum I = 0$

Here, $\sum I$ is the sum of currents meet at junction.

According to Kirchhoff’s junction rule, the sum of currents at junction is equal to zero.

${I_1} + {I_2} + {I_3} + {I_4} = 0$

Here, ${I_1}$ is the current in first wire, ${I_2}$ is the current in second wire, ${I_3}$ is the current in third wire, and ${I_4}$ is the current in fourth wire.

Substitute 1.47 A for ${I_1},$ 2.75 A for ${I_2}$ and $- 6.35\;{\rm{A}}$ for ${I_3}$ in the above equation to solve for ${I_4}.$

$\begin{array}{c}\\1.47\,{\rm{A}} + 2.75\,{\rm{A}} - 6.35\,{\rm{A}} + {I_4} = 0\\\\{I_4} = 6.35\,{\rm{A}} - 1.47\,{\rm{A}} - 2.75\,{\rm{A}}\\\\{I_4} = 2.13\,{\rm{A}}\\\end{array}$

The direction of current in the fourth wire is into the junction because it is positive.

Ans:

The current in the fourth wire is 2.13 A.

The direction of current in the fourth wire is into the junction.