Question

Three resistors are connected to form a T shape network as shown in the figure. Calculate the current following through the $8\mathrm{K}$ resistor?

Answer 1

Step 1: Kirchoff's current law

1. According to the Kirchhoff current law (KCL), the total current entering a junction or a node is equal to the current leaving the node as no charge is lost.
2. A node refers to a junction connecting two or more current-carrying routes like cables and other components.
3. Kirchhoff’s current law can also be applied to analyze parallel circuits.

Step 2: Apply Kirchoff's current law

Let, the potential at the junction be $\mathrm{x}$.

Step 3: Calculation

Using Ohm's law to calculate the individual current following at the junction.

$$\begin{gathered} V=IR \\ \Rightarrow I=\frac{V}{R} \end{gathered}$$

Where, $I$ is the current flowing through the circuit, $V$ is the voltage, and, $R$ is the resistance.

By applying KCL at the junction.

The net current at $\mathrm{x}$ will be:

${\mathrm{I}}_{\mathrm{entering}}+(-{\mathrm{I}}_{\mathrm{existing}})=0$

then,

$I_1+I_2+I_3=0$

$\frac{\mathrm{x}-2}{2}+\frac{\mathrm{x}-(-8)}{4}+\frac{\mathrm{x}-(-4)}{8}=0$

By taking L.C.M, the potential at the junction is

$\mathrm{x}=-\frac{12}{7}\mathrm{V}$

Step 4: Calculate the current flowing through the resistor $8\mathrm{K}$

Using the expression below to find the current flowing through the resistor $8\mathrm{K}$

$I=\frac{V}{R}$

$$\begin{gathered} \mathrm{I}=\frac{\left(-{\displaystyle \frac{12}{7}}+4\right)}{8} \\ =0.285\mathrm{A} \end{gathered}$$

Therefore, the current flow in the $8\mathrm{K}$ resistor is $0.285\mathrm{A}$.