Parallel Resistors Continued
In contrast to the series case, when we add resistors in parallel, we create more paths along which current can flow. By doing this we decrease the total resistance of the circuit!
Take a look at the diagram below. On the left we have the same circuit as in the previous section with a battery and a resistor. The ammeter shows a current of \(\text{1}\) \(\text{A}\). On the right we have added a second resistor in parallel to the first resistor. This has increased the number of paths (branches) the charge can take through the circuit - the total resistance has decreased. You can see that the current in the circuit has increased. Also notice that the current in the different branches can be different.
The total resistance of a number of parallel resistors is NOT the sum of the individual resistances as the overall resistance decreases with more paths for the current. The total resistance for parallel resistors is given by:
\[\frac{1}{{R}_{P}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\ldots\]Let us consider the case where we have two resistors in parallel and work out what the final resistance would be. This situation is shown in the diagram below:
Applying the formula for the total resistance we have:
\begin{align*} \frac{1}{{R}_{P}} & = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \ldots \\ & \text{There are only two resistors} \\ \frac{1}{{R}_{P}} & = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} \\ & \text{Add the fractions} \\ \frac{1}{{R}_{P}} & = \frac{1}{{R}_{1}} \times \frac{{R}_{2}}{{R}_{2}} + \frac{1}{{R}_{2}} \times \frac{{R}_{1}}{{R}_{1}} \\ \frac{1}{{R}_{P}} & = \frac{{R}_{2}}{{R}_{1}{R}_{2}} + \frac{{R}_{1}}{{R}_{1}{R}_{2}} \\ & \text{Rearrange} \\ \frac{1}{{R}_{P}} & = \frac{{R}_{2} + {R}_{1}}{{R}_{1}{R}_{2}} \\ \frac{1}{{R}_{P}} & = \frac{{R}_{1} + {R}_{2}}{{R}_{1}{R}_{2}} \\ {R}_{P} & = \frac{{R}_{1}{R}_{2}}{{R}_{1} + {R}_{2}} \end{align*}For any two resistors in parallel we now know that
\[{R}_{P} = \frac{\text{product of resistances}}{\text{sum of resistances}} = \frac{{R}_{1}{R}_{2}}{{R}_{1} + {R}_{2}}\]Optional Experiment: Current Dividers
Aim
Test what happens to the current and the voltage in series circuits when additional resistors are added.
Apparatus
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A battery
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A voltmeter
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An ammeter
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Wires
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Resistors
Method
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Connect each circuit shown below
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Measure the voltage across each resistor in the circuit.
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Measure the current before and after each resistor in the circuit and before and after the parallel branches.
Results and conclusions
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Compare the currents through individual resistors with each other.
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Compare the sum of the currents through individual resistors with the current before the parallel branches.
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Compare the various voltage measurements across the parallel resistors.
Example: Parallel Resistors I
Question
A circuit contains two resistors in parallel. The resistors have resistance values of \(\text{15}\) \(\text{Ω}\) and \(\text{7}\) \(\text{Ω}\).
What is the total resistance in the circuit?
Step 1: Analyze the question
We are told that the resistors in the circuit are in parallel circuit and that we need to calculate the total resistance. The values of the two resistors have been given in the correct units, Ω.
Step 2: Apply the relevant principles
The total resistance for resistors in parallel has been shown to be the product of the resistances divided by the sum. We can use
\[{R}_{P} = \frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}\]We have only two resistors and we now the resistances. In this case we have that:
\begin{align*} {R}_{P} & = \frac{{R}_{1}{R}_{2}}{{R}_{1} + {R}_{2}} \\ & = \frac{\left(\text{15}\text{ Ω} \right)\left(\text{7}\text{ Ω} \right)}{\text{15}\text{ Ω} + \text{7}\text{ Ω}} \\ & = \frac{105 {\Omega }^{2}}{\text{22}\text{ Ω}} \\ & = \text{4.77}\text{ Ω} \end{align*}Step 3: Quote the final result
The total resistance of the resistors in parallel is \(\text{4.77}\) \(\text{Ω}\)
Example: Parallel Resistors II
Question
We add a third parallel resistor to the configuration (setup) in the previous example. The additional resistor has a resistance of \(\text{3}\) \(\text{Ω}\).
What is the total resistance in the circuit?
Step 1: Analyze the question
We are told that the resistors in the circuit are in parallel circuit and that we need to calculate the total resistance. The value of the additional resistor has been given in the correct units, Ω.
Step 2: Apply the relevant principles
The total resistance for resistors in parallel has been given as
\[\frac{1}{{R}_{P}} = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \ldots\]We have three resistors and we now the resistances. In this case we have that:
\begin{align*} \frac{1}{{R}_{P}} & = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \ldots \\ & \text{there are three resistors} \\ \frac{1}{{R}_{P}} & = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \frac{1}{{R}_{3}} \\ & \text{add the fractions} \\ \frac{1}{{R}_{P}} & = \frac{1}{{R}_{1}} \times \frac{{R}_{2}{R}_{3}}{{R}_{2}{R}_{3}} + \frac{1}{{R}_{2}} \times \frac{{R}_{1}{R}_{3}}{{R}_{1}{R}_{3}} + \frac{1}{{R}_{3}} \times \frac{{R}_{1}{R}_{2}}{{R}_{1}{R}_{2}} \\ \frac{1}{{R}_{P}} & = \frac{{R}_{2}{R}_{3}}{{R}_{1}{R}_{2}{R}_{3}} + \frac{{R}_{1}{R}_{3}}{{R}_{1}{R}_{2}{R}_{3}} + \frac{{R}_{1}{R}_{2}}{{R}_{1}{R}_{2}{R}_{3}} \\ & \text{rearrange} \\ \frac{1}{{R}_{P}} & = \frac{{R}_{2}{R}_{3} + {R}_{1}{R}_{3} + {R}_{2}{R}_{3}}{{R}_{1}{R}_{2}{R}_{3}} \\ {R}_{P} & = \frac{{R}_{1}{R}_{2}{R}_{3}}{{R}_{2}{R}_{3} + {R}_{1}{R}_{3} + {R}_{2}{R}_{3}} \\ {R}_{P} & = \frac{\left(\text{15}\text{ Ω} \right)\left(\text{7}\text{ Ω} \right)\left(\text{3}\text{ Ω} \right)}{\left(\text{7}\text{ Ω} \right)\left(\text{3}\text{ Ω} \right) + \left(\text{15}\text{ Ω} \right)\left(\text{3}\text{ Ω}\right) + \left(\text{7}\text{ Ω} \right)\left(\text{15}\text{ Ω}\right)} \\ {R}_{P} & = \frac{315 {\Omega }^{3}}{21 {\Omega }^{2} + 45 {\Omega }^{2} + 105 {\Omega }^{2}} \\ {R}_{P} & = \frac{315 {\Omega }^{3}}{171 {\Omega }^{2}} \\ {R}_{P} & = \text{1.84}\text{ Ω} \end{align*}Step 3: Quote the final result
The total resistance of the resistors in parallel is \(\text{1.84}\) \(\text{Ω}\)
When calculating the resistance for complex resistor configurations, you can start with any combination of two resistors (in series or parallel) and calculate their total resistance. Then you can replace them with a single resistor that has the total resistance you calculated. Now use this new resistor in combination with any other resistor and repeat the process until there is only one resistor left. In the above example we could just have used the answer from the first example in parallel with the new resistor and we would get the same answer.
Example: Parallel Resistors III
Question
We add a third parallel resistor to the first parallel worked example configuration (setup). The additional resistor has a resistance of \(\text{3}\) \(\text{Ω}\)
What is the total resistance in the circuit?
Step 1: Analyze the question
We are told that the resistors in the circuit are in parallel circuit and that we need to calculate the total resistance. The value of the additional resistor has been given in the correct units, Ω.
Step 2: Apply the relevant principles
We can swap the resistors around without changing the circuit:
We have already calculated the total resistance of the two resistors in the dashed box to be \(\text{4.77}\) \(\text{Ω}\). We can replace these two resistors with a single resistor of \(\text{4.77}\) \(\text{Ω}\) to get:
Step 3: Calculate the total resistance for the next pair of resistors
Then we use the formula for two parallel resistors again to get the total resistance for this new circuit:
\begin{align*} {R}_{P} & = \frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}} \\ & = \frac{\left(\text{4.77}\text{ Ω} \right)\left(\text{3}\text{ Ω}\right)}{\text{4.77}\text{ Ω} + \text{3}\text{ Ω} } \\ & = \frac{\text{14.31} {\Omega }^{2}}{\text{11.77} \Omega } \\ & = \text{1.84}\text{ Ω} \end{align*}Step 4: Quote the final result
The total resistance of the resistors in parallel is \(\text{1.84}\) \(\text{Ω}\). This is the same result as when we added all three resistors together at once.
This lesson is part of:
Electric Circuits