This problem aims to familiarize us with different **circuital laws** and **circuit analysis.** The concepts required to solve this problem are related to **Kirchoff’s circuit laws,** which include** Kirchoff’s first law,** known as the **current law,** and **Kirchoff’s second law,** known as the **voltage law**.

In circuit analysis, **Kirchhoff’s circuit laws** help to form an equation for respective components such as a **resistor, capacitor, or inductor**. Now according to **Kirchoff’s first law**, the total **charge** entering a junction (also known as a node) is **equal** to the total **charge** exiting the junction since no charge is wasted.

Let’s say the **currents** $I_1, I_2$ and $I_3$ are **entering** the node, so taking them as **positive,** and the currents $I_4$ and $I_5$ are **exiting** the nodes, thus **negative.** This forms an **equation** according to the statement:

\[I_1 + I_2 + I_3 – I_4 – I_5=0\]

According to **Kirchoff’s second law**, the voltage of a **closed** loop is equal to the sum of every **potential** decline in that loop, which equals **zero.**

\[V_{AB}+V_{BC}+V_{CD}+V_{DA}=0\]

## Expert Answer

To start the solution, we will be using **Kirchhoff’s loop rule**. We shall start by drawing a **current** via each **resistor.** This step basically shows the **directions** preferred for the **currents.** These chosen **directions** are **random,** and if found to be incorrect, then the **negative** value of the calculated **current** will indicate that the analysis was the **opposite.**

Now let’s **mark** both ends of every **resistor** with $+$ and $-$ that help in identifying the **voltage drops** and **peaks.** We know that the direction of **conventional current** is always from a higher potential to a lower potential.

Applying **Kirchoff’s voltage rule** to the loop $ABCF$:

\[V_1+I_2R_2=I_1R_1\]

Similarly, for the other **loop** $FCDE$:

\[V_2=I_2R_2\]

Solving this **equation** for $I_2$ gives us:

\[I_2=\dfrac{V_2}{R_2}\]

\[=\dfrac{12 V}{2.0\Omega}\]

\[I_2=6.0\space A\]

Since $I_2$ is a **positive value,** the current in $R_2$ goes as shown in the figure. Now solving the first **equation** for $I_1$:

\[I_1=\dfrac{V_1+I_2R_2}{R_1}\]

Substituting $I_2=V_2/R_2$:

\[I_1=\dfrac{V_1+\dfrac{V_2}{R_2}R_2}{R_1}\]

\[I_1=\dfrac{V_1+V_2}{R_1}\]

\[I_1=\dfrac{4.0 V+12 V}{8.0}\]

\[I_1=2.0\space A\]

Since $I_1$ also comes out to be a **positive value,** the **current** in the resistor $R_1$ goes as shown in the figure.

## Numerical Result

$I_2=6.0\space A$ is a **positive value,** and the **current** in the resistor $R_2$ goes from **left to right**.

$I_1= 2.0\space A$ also comes out to be a **positive value,** so the **current** in the resistor $R_1$ goes from **left to right.**

## Example

A $60.0\Omega$ resistor is in **parallel** with a $120\Omega$ resistor. This **parallel connection** is in **series** with a $20.2\Omega$ resistor **connected** across a $15.0 V$ battery. Find the **current** and the **power** supplied to the $120\Omega$.

The **current** in the $120.0\Omega$ resistor is $I_{120} = \dfrac{V_{AB}}{120.0}$, but the **equivalent resistance** $R_{AB}$ is:

\[\dfrac{1}{R_{AB}}=\dfrac{1}{60.0}+\dfrac{1}{120.0} = 40.0\Omega\]

This **resistance** of $40.0\Omega$ is in **series** with the $20.0\Omega$, thus total **Resistance** is $40.0\Omega+20.0\Omega=60.0\Omega$. Using **ohm’s law,** the total current from the **battery** is:

\[I=\dfrac{15.0V}{60.0\Omega}=0.250\space A\]

Now for $V_{AB}$:

\[V_{AB}=(0.250A)R_{AB}=0.250\times40.0=10.0\space V\]

Finally, the **current** from $120.0\Omega$ is:

\[I_{120}=\dfrac{10.0}{120.0}=8.33\times 10^{-2}\space A\]

And the **power** delivered is:

\[P=I_{120}^{2}R=(8.33\times 10^{-2})^2(120.0)=0.833\space W\]

*Images/Mathematical drawings are created with Geogebra.*