## Use of Ohm’s Law in Parallel Circuits

Contents

- Use of Ohm’s Law in Parallel Circuits
- Example: Ohm’s Law, Parallel Circuit
- Example: Ohm’s Law, Parallel Circuit
- Example: Ohm’s Law, Parallel Circuit
- Question
- Step 1: First draw the circuit before doing any calculations
- Step 2: Determine how to approach the problem
- Step 3: Calculate the current through the cell
- Step 4: Now determine the current through one of the parallel resistors
- Step 5: Calculate the current through the other parallel resistor
- Step 6: Write the final answer

Using the definitions for equivalent resistance for resistors in parallel, we can analyze some circuits with these setups.

### Parallel circuits

Consider a circuit consisting of a single cell and three resistors that are connected in parallel.

The first principle to understand about parallel circuits is that the voltage is equal across all components in the circuit. This is because there are only two sets of electrically common points in a parallel circuit, and voltage measured between sets of common points must always be the same at any given time. So, for the circuit shown, the following is true:

\[V = V_{1} = V_{2} = V_{3}.\]

The second principle for a parallel circuit is that all the currents through each resistor must add up to the total current in the circuit:

\[I = I_{1} + I_{2} + I_{3}.\]

Using these principles and our knowledge of how to calculate the equivalent resistance of parallel resistors, we can now approach some circuit problems involving parallel resistors.

## Example: Ohm’s Law, Parallel Circuit

### Question

Calculate the current (I) in this circuit if the resistors are both ohmic in nature.

### Step 1: Determine what is required

We are required to calculate the current flowing in the circuit.

### Step 2: Determine how to approach the problem

Since the resistors are ohmic in nature, we can use Ohm’s Law. There are however two resistors in the circuit and we need to find the total resistance.

### Step 3: Find the equivalent resistance in circuit

Since the resistors are connected in parallel, the total (equivalent) resistance R is:

\[\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}.\]\begin{align*} \frac{1}{R} &= \frac{1}{R_1} + \frac{1}{R_2} \\ &= \frac{1}{2} + \frac{1}{4} \\ &= \frac{2+1}{4} \\ &= \frac{3}{4} \\ \text{Therefore. } R &= \text{1.33} \Omega\end{align*}

### Step 4: Apply Ohm’s Law

\begin{align*} R&= \frac{V}{I} \\ R \cdot \frac{I}{R} &= \frac{V}{I} \cdot \frac{I}{R} \\ I &= \frac{V}{R}\\ I &= V \cdot \frac{1}{R}\\ &= (12) \left(\frac{3}{4}\right) \\ &= \text{9}\text{ A}\end{align*}

### Step 5: Write the final answer

The current flowing in the circuit is \(\text{9}\) \(\text{A}\).

## Example: Ohm’s Law, Parallel Circuit

### Question

Two ohmic resistors (\(R_1\) and \(R_2\)) are connected in parallel with a cell. Find the resistance of \(R_2\), given that the current flowing through the cell is \(\text{4.8}\) \(\text{A}\) and that the voltage across the cell is \(\text{9}\) \(\text{V}\).

### Step 1: Determine what is required

We need to calculate the resistance \(R_2\).

### Step 2: Determine how to approach the problem

Since the resistors are ohmic and we are given the voltage across the cell and the current through the cell, we can use Ohm’s Law to find the equivalent resistance in the circuit.\begin{align*}R & = \frac{V}{I} \\& = \frac{9}{\text{4.8}} \\& = \text{1.875} \ \Omega\end{align*}

### Step 3: Calculate the value for \(R_2\)

Since we know the equivalent resistance and the resistance of \(R_1\), we can use the formula for resistors in parallel to find the resistance of \(R_2\).\begin{align*}\frac{1}{R} & = \frac{1}{R_1} + \frac{1}{R_2}\end{align*}Rearranging to solve for \(R_2\):\begin{align*}\frac{1}{R_2} & = \frac{1}{R} – \frac{1}{R_1} \\& = \frac{1}{\text{1.875}} – \frac{1}{3}\\& = \text{0.2} \\R_2 & = \frac{1}{\text{0.2}} \\& = \text{5} \ \Omega\end{align*}

### Step 4: Write the final answer

The resistance \(R_2\) is \(\text{5}\) \(\Omega\)

## Example: Ohm’s Law, Parallel Circuit

### Question

An 18 volt cell is connected to two parallel resistors of \(\text{4}\) \(\Omega\) and \(\text{12}\) \(\Omega\) respectively. Calculate the current through the cell and through each of the resistors.

### Step 1: First draw the circuit before doing any calculations

### Step 2: Determine how to approach the problem

We need to determine the current through the cell and each of the parallel resistors. We have been given the potential difference across the cell and the resistances of the resistors, so we can use Ohm’s Law to calculate the current.

### Step 3: Calculate the current through the cell

To calculate the current through the cell we first need to determine the equivalent resistance of the rest of the circuit. The resistors are in parallel and therefore:\begin{align*}\frac{1}{R} &= \frac{1}{R_1} + \frac{1}{R_2} \\&= \frac{1}{4} + \frac{1}{12} \\&= \frac{3+1}{12} \\&= \frac{4}{12} \\R &= \frac{12}{4} = \text{3} \ \Omega\end{align*}Now using Ohm’s Law to find the current through the cell:\begin{align*}R &= \frac{V}{I} \\I &= \frac{V}{R} \\&= \frac{18}{3} \\I &= \text{6}\text{ A}\end{align*}

### Step 4: Now determine the current through one of the parallel resistors

We know that for a purely parallel circuit, the voltage across the cell is the same as the voltage across each of the parallel resistors. For this circuit:\begin{align*}V &= V_1 = V_2 = \text{18}\text{ V}\end{align*}Let’s start with calculating the current through \(R_1\) using Ohm’s Law:\begin{align*}R_1 &= \frac{V_1}{I_1} \\I_1 &= \frac{V_1}{R_1} \\&= \frac{18}{4} \\I_1 &= \text{4.5}\text{ A}\end{align*}

### Step 5: Calculate the current through the other parallel resistor

We can use Ohm’s Law again to find the current in \(R_2\):\begin{align*}R_2 &= \frac{V_2}{I_2} \\I_2 &= \frac{V_2}{R_2} \\&= \frac{18}{12} \\I_2 &= \text{1.5}\text{ A}\end{align*}An alternative method of calculating \(I_2\) would have been to use the fact that the currents through each of the parallel resistors must add up to the total current through the cell:\begin{align*}I &= I_1 + I_2 \\I_2 &= I – I_1 \\&= 6 – 4.5 \\I_2 &= \text{1.5}\text{ A}\end{align*}

### Step 6: Write the final answer

The current through the cell is \(\text{6}\) \(\text{A}\).

The current through the \(\text{4}\) \(\Omega\) resistor is \(\text{4.5}\) \(\text{A}\).

The current through the \(\text{12}\) \(\Omega\) resistor is \(\text{1.5}\) \(\text{A}\).