The main objective of this question is to find the **direction and magnitude** of the **current** in **0.2 ohm** and **0.8 ohm** resistors.

This question uses the concept of **Kirchoff’s current law and Kirchhoff’s voltage law** to find the **direction and magnitude of current** for the given circuit diagram. In **Kirchoff’s current law**, the **current entering** the node must be **equal** to the **current leaving the node** while in **Kirchoff’s voltage** **law** the **overall sum** of **voltage** is equal to **zero**.

## Expert Answer

We are** given** with:

$ V_1 =4.0 v $

$ R_1=8.0 ohm$

$ V_2=12v$

$R_2=2.0 ohm $

We have to find the **direction and magnitude** of the current in the $8.0$ ohm and $2.0$ ohm resistor.

So, **applying Kirchoff’s current law** which is:

\[i_1 \space – \space i_2 \space – \space i_3 \]

\[4 \space – \space 8i_3 \space + \space 2i_2 \space = \space 0 \]

Now **applying Kirchoff’s voltage** law results in:

\[\space -2i_2 \space + \space 12 \space = \space 0 \]

**Then**:

\[2i_2 \space = \space 12\]

**Dividing** by $2$ will result in:

\[i_2 \space = \space 6 \space a \pm \]

**Putting** the **value** of $i_2$ results in:

\[4 \space – \space 8i_3 \space + \space 2 \space \times\ 6 \space = \space 0 \]

\[16 \space – \space 8i_3 \space = \space 0\]

\[8i_3 \space = \space 16 \]

\[i_3 \space = \space 2a \space \pm \]

So, **putting the value** of $i_3$ will result in:

\[i_1 \space = \space i_2 \space + \space i_3 \space = \space 8a \pm\]

**Thus** $i_1$ is equal to $8a$ \pm.

## Numeric Answer

The **current** $i_1$ is $8a$ \pm while the **current** $i_2$ is $6a$ \pm and **current** $i_3$ is $2a$ \pm .

## Example

In this question, you are required to find the direction and magnitude of the current in $10$ ohm and $4$ ohm resistors and the voltage $V_1$ is $4.0 v$ and the $V_2$ is $12v$.

We are **given** the** following** **data**:

$V_1 =4.0 v$.

$R_1=10.0 ohm$.

$V_2=12v$.

$R_2=4.0 ohm$.

In this question, we have to find the **direction and magnitude** of the **current** in the $10.0$ ohm and $4.0$ ohm resistor.

So, **applying Kirchoff’s current law** which is **mathematically** represented as:

\[i_1 \space – \space i_2 \space – \space i_3 \]

\[4 \space – \space 10i_3 \space + \space 2i_2 \space = \space 0 \]

Now **applying Kirchoff’s voltage law** which is mathematically represented as:

\[\space -4i_2 \space + \space 12 \space = \space 0 \]

Then:

\[4i_2 \space = \space 12\]

**Dividing** by 4 will result in:

\[i_2 \space = \space 3 \space a \pm \]

**Putting** the value of $i_2$ results in:

\[4 \space – \space 10i_3 \space + \space 2 \space \times\ 3 \space = \space 0 \]

\[10 \space – \space 8i_3 \space = \space 0\]

\[8i_3 \space = \space 10 \]

\[i_3 \space = \space 1.25a \space \pm \]

So, **putting the value** of $i_3$ will result in:

\[i_1 \space = \space i_2 \space + \space i_3 \space = \space 4.25a \pm\]

Hence, the **current** in the $10-ohm$ and $4-ohm$ resistor is $1.25-ohm$ and $3-ohm$, **respectively**.