Question Video: Finding the Equivalent Resistance of a Combination Circuit | Nagwa Question Video: Finding the Equivalent Resistance of a Combination Circuit | Nagwa

Question Video: Finding the Equivalent Resistance of a Combination Circuit Physics • Third Year of Secondary School

The circuit shown contains both series and parallel combinations of resistors. What is the total current in the circuit shown? Give your answer to one decimal place. What is the total power dissipated by the circuit? Give your answer to one decimal place.

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Video Transcript

The circuit shown contains both series and parallel combinations of resistors. What is the total current in the circuit shown? Give your answer to one decimal place.

Considering our circuit, we see that it involves four resistors and one cell right here. The positive terminal of that cell points to the right. So, conventionally, charge will flow in a clockwise direction around the circuit. After that charge passes through the first resistor, it will get to this branch point right here. The current will divide up based on the relative resistances of these two parallel branches, travel through the branches, and then rejoin. After this, the total or combined current in the circuit will then pass through this 1.6-ohm resistor and on to the negative terminal of our cell.

Let’s call the total current in our circuit 𝐼 sub t. It’s that value that we want to solve for. We can begin doing this by recalling Ohm’s law, which tells us that the potential difference across a circuit equals the current in that circuit multiplied by the total circuit resistance. Dividing both sides of Ohm’s law by the resistance 𝑅, we find that 𝐼 is equal to 𝑉 divided by 𝑅. So then, if we know the total potential difference across our circuit 𝑉 and the total resistance of the circuit 𝑅, we can use these values to solve for the total circuit current 𝐼. In our case, the total circuit current 𝐼 sub t is equal to 5.5 volts, the potential difference supplied by our cell, divided by the total resistance in the circuit that we’ll call 𝑅 sub t.

To solve for 𝑅 sub t, we’ll need to combine the resistances of all four of our resistors into one effective resistance. This process will take a few steps because as our problem statement notes, here we have resistors connected both in series and in parallel. Let’s begin by considering the resistors that are arranged in parallel. In general, if we have exactly two resistors, 𝑅 one and 𝑅 two, in parallel with one another, then the effective resistance of these resistors, we’ll call it 𝑅 sub p, is 𝑅 one times 𝑅 two divided by 𝑅 one plus 𝑅 two.

As we combine resistances in our circuit to arrive at one total resistance for the circuit, we can say that effectively this 2.5-ohm resistor and the 3.2-ohm resistor in parallel is the same as having one single resistor with a resistance value of 2.5 ohms times 3.2 ohms divided by 2.5 ohms plus 3.2 ohms. Here, we’ve applied our rule for the equivalent resistance of two resistors in parallel.

Note that we now effectively have three resistors in our circuit. And these three resistors are all in series. The equivalent resistance of three resistors arranged in series with one another is equal to the sum of the individual resistance values. All this means we now have enough information to write an expression for the total resistance in our circuit 𝑅 sub t. By our series resistor addition rule, we know that 𝑅 sub t equals the resistance of our first resistor, 1.5 ohms, plus the equivalent resistance of our second resistor plus that of our third resistor, 1.6 ohms. If we enter this expression for 𝑅 sub t onto our calculator, we get a result of about 4.503 ohms.

We now have a value we can use for 𝑅 sub t in our expression to solve for the total circuit current. 𝐼 sub t equals 5.5 volts divided by 4.503 and so on ohms. And to one decimal place, this is equal to 1.2 amperes. To this level of precision, 1.2 amperes is the total current in the circuit.

Let’s now look at part two of our question.

What is the total power dissipated by the circuit? Give your answer to one decimal place.

An electrical circuit dissipates power by transferring energy away from the circuit over time. For example, power is dissipated when resistors heat up and transfer heat energy to their surroundings. This part of our question asks us about the total power dissipated by the circuit. That power is equal in general to the potential difference across the circuit multiplied by the total circuit current.

In our case, that potential difference is 5.5 volts and that current, as we’ve seen, is 5.5 volts divided by the total circuit resistance. We see that this is equal to the potential difference provided by the cell squared divided by the total circuit resistance. Calculating this fraction to one decimal place, we get a result of 6.7 with units of watts. This is how many joules of energy are transferred away from our circuit every second. In other words, the power dissipated by the circuit is 6.7 watts.

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