Ohm's Law in Series Circuits
Use of Ohm's Law in Series Circuits
Using the definitions for equivalent resistance for resistors in series, we can analyze some circuits with these setups.
Series circuits
Consider a circuit consisting of three resistors and a single cell connected in series.
The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a series circuit. From the way that the battery is connected, we can tell in which direction the current will flow. We know that current flows from positive to negative by convention. Conventional current in this circuit will flow in a clockwise direction, from point A to B to C to D and back to A.
We know that in a series circuit the current has to be the same in all components. So we can write:
\[I = I_{1} = I_{2} = I_{3}.\]We also know that total voltage of the circuit has to be equal to the sum of the voltages over all three resistors. So we can write:
\[V = V_{1} + V_{2} + V_{3}\]Using this information and what we know about calculating the equivalent resistance of resistors in series, we can approach some circuit problems.
Example: Ohm's Law, Series 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 total resistance in circuit
Since the resistors are connected in series, the total (equivalent) resistance R is:
\[R = R_{1} + R_{2}\]Therefore,
\begin{align*} R & = \text{2} + \text{4} \\ & = \text{6}\text{ Ω} \end{align*}Step 4: Apply Ohm's Law
\begin{align*} R & = \frac{V}{I} \\ R \times \frac{I}{R} & = \frac{V}{I} \times \frac{I}{R} \\ I & = \frac{V}{R} \\ & = \frac{12}{6} \\ & = \text{2}\text{ A} \end{align*}
Step 5: Write the final answer
A current of \(\text{2}\) \(\text{A}\) is flowing in the circuit.
Example: Ohm's Law, Series Circuit
Question
Two ohmic resistors (\(R_{1}\) and \(R_{2}\)) are connected in series with a cell. Find the resistance of \(R_{2}\), given that the current flowing through \(R_{1}\) and \(R_{2}\) is \(\text{0.25}\) \(\text{A}\) and that the voltage across the cell is \(\text{1.5}\) \(\text{V}\). \(R_{1}\) =\(\text{1}\) \(\text{Ω}\).
Step 1: Draw the circuit and fill in all known values.
Step 2: Determine how to approach the problem.
We can use Ohm's Law to find the total resistance R in the circuit, and then calculate the unknown resistance using:
\[R = R_{1} + R_{2}\]because it is in a series circuit.
Step 3: Find the total resistance
\begin{align*} R & = \frac{V}{I} \\ & = \frac{\text{1.5}}{\text{0.25}} \\ & = \text{6}\text{ Ω} \end{align*}Step 4: Find the unknown resistance
We know that:
\[R = \text{6}\text{ Ω}\]and that
\[R_{1} = \text{1}\text{ Ω}\]Since
\[R = R_{1} + R_{2}\]\[R_{2} = R - R_{1}\]Therefore,
\[R_{1} = \text{5}\text{ Ω}\]Example: Ohm's Law, Series Circuit
Question
For the following circuit, calculate:
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the voltage drops \(V_1\), \(V_2\) and \(V_3\) across the resistors \(R_1\), \(R_2\), and \(R_3\)
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the resistance of \(R_3\).
Step 1: Determine how to approach the problem
We are given the voltage across the cell and the current in the circuit, as well as the resistances of two of the three resistors. We can use Ohm's Law to calculate the voltage drop across the known resistors. Since the resistors are in a series circuit the voltage is \(V = V_1 + V_2 + V_3\) and we can calculate \(V_3\). Now we can use this information to find the voltage across the unknown resistor \(R_3\).
Step 2: Calculate voltage drop across \(R_1\)
Using Ohm's Law:\begin{align*} R_1 &= \frac{V_1}{I} \\ I \cdot R_1 &= I \cdot \frac{V_1}{I} \\ V_1 &= {I}\cdot{R_1}\\ &= 2 \cdot 1 \\ V_1 &= \text{2}\text{ V} \end{align*}
Step 3: Calculate voltage drop across \(R_2\)
Again using Ohm's Law:\begin{align*} R_2 &= \frac{V_2}{I} \\ I \cdot R_2 &= I \cdot \frac{V_2}{I} \\ V_2 &= {I}\cdot{R_2}\\ &= 2 \cdot 3 \\ V_2 &= \text{6}\text{ V} \end{align*}
Step 4: Calculate voltage drop across \(R_3\)
Since the voltage drop across all the resistors combined must be the same as the voltage drop across the cell in a series circuit, we can find \(V_3\) using:\begin{align*} V &= V_1 + V_2 + V_3\\ V_3 &= V - V_1 - V_2 \\ &= 18 - 2 - 6 \\ V_3&= \text{10}\text{ V} \end{align*}
Step 5: Find the resistance of \(R_3\)
We know the voltage across \(R_3\) and the current through it, so we can use Ohm's Law to calculate the value for the resistance:\begin{align*} R_3 &= \frac{V_3}{I}\\ &= \frac{10}{2} \\R_3&= \text{5} \Omega\end{align*}
Step 6: Write the final answer
\(V_1 = \text{2}\text{ V}\)
\(V_2 = \text{6}\text{ V}\)
\(V_3 = \text{10}\text{ V}\)
\(R_1 = \text{5} \Omega\)
This lesson is part of:
Electric Circuits