Question Video: Analyzing Parallel Circuits | Nagwa Question Video: Analyzing Parallel Circuits | Nagwa

# Question Video: Analyzing Parallel Circuits Physics • Third Year of Secondary School

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In the circuit shown, the current takes multiple paths from the positive battery terminal to the negative battery terminal. Find the total resistance of the circuit.

03:04

### Video Transcript

In the circuit shown, the current takes multiple paths from the positive battery terminal to the negative battery terminal. Find the total resistance of the circuit.

That given diagram shows a combination circuit as resistors are both in parallel as well as in series. Before we simplify our circuit to solve for the total resistance, letβs remind ourselves how to find total resistance in both series and parallel circuits. In a series circuit, the total resistance or equivalent resistance is equal to the sum of each individual resistor, π one, π two, π three, and so on, until all the resistors are accounted for. In a parallel circuit, one over the total resistance or equivalent resistance is equal to one over π one plus one over π two plus one over π three and so on and so forth until all the resistors are accounted for.

Looking at our diagram, we can see that if we could replace our two resistors in parallel with an equivalent resistor, then we would have a circuit that is a simple series circuit. We have drawn a simplified version of the circuit below. We need to determine what value of π will be equivalent to the 12-ohm and 18-ohm resistors in parallel. To do this, we need to use the equation to find the total resistance of a parallel circuit.

We used π for equivalent resistance, so the left side of the equation becomes one over π. The right side of the equation are our two resistors, one over 12 ohms plus one over 18 ohms. When adding fractions, we need to find the least common denominator. The least common denominator for 12 ohms and 18 ohms would be 36 ohms. In order for each of our fractions to have a denominator of 36 ohms, our one-over-12-ohm fraction needs to be multiplied by three over three and our one-over-18-ohm fraction has to be multiplied by two over two. This makes our fractions three over 36 ohms plus two over 36 ohms. When we add three over 36 ohms plus two over 36 ohms, we get five over 36 ohms.

To isolate π, we multiply both sides of the equation by 36 ohms and π. This cancels out π on the left side of the equation and 36 ohms on the right side of the equation, giving us the equation 36 ohms is equal to five π. Then, we divide both sides by five, canceling out the five on the right side of the equation, giving us an equivalent resistance of 36 divided by five ohms. In decimal form, this would be 7.2 ohms.

Looking at our simplified circuit, we can see that our three resistors are in series with each other. Therefore, we need to use the series equation to find total resistance. The total resistance of our circuit, π equivalent, is equal to 14 ohms, the value of the first resistor, plus 10 ohms, the value of the second resistor, plus 7.2 ohms, the value of the equivalent resistance of the 12- and 18-ohm resistors that were in parallel. When we add our three resistors together, we get a value of 31.2 ohms. All the values in our problem have been given to two significant figures. Therefore, we need to round our total resistance to two significant figures, which gives us a total resistance for our circuit of 31 ohms.

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