Video: Calculating Currents and Voltages in a Five-Resistor Circuit

The total current in the circuit below is 100 mA. a) Calculate, to the nearest ohm, the total resistance of the parallel combination of resistors. b) What is the total resistance in this circuit to the nearest ohm? c) What is the voltage across the 500 Ω resistor to 2 significant figures? d) What is the current in the 600 Ω resistor to the nearest milliampere?

07:35

Video Transcript

The total current below is 100 milliamperes. Calculate, to the nearest ohm, the total resistance of the parallel combination of resistors.

When a circuit is in parallel, it means that there are two or more paths for the current to flow. So the portion of the circuit that we’re interested in with the parallel combination of resistors is this part of the circuit, because the current can flow either over the path with the 200-ohm resister or the path with the 600- and the 500-ohm resistor. So before we find the total resistance of this portion of the circuit, like the question asks, let’s review how we can combine resistors in parallel and in series.

If we have a circuit with resistors in series, that is, the resistors are connected in line with only one path for the current to flow, we can find the total resistance by simply adding the resistances of each resistor. If we have resistors that are in parallel, we can find their total resistance by summing the inverse of their resistances and then taking the inverse of that. Now, let’s find the total resistance of this parallel portion of our circuit. First, let’s find the total resistance of the bottom two resistors. Since they’re in series, we can find their total resistance by adding their resistances together. So the total resistance for these two resistors is 600 ohms plus 500 ohms, which gives us 1100 ohms. We can now draw an equivalent circuit with the 1100-ohm resistor instead of the 600- and 500-ohm resistors.

Now, to find the total resistance of this portion of the circuit, we need to combine the 200-ohm resistor and the 1100-ohm resistor. We can now find the total resistance of this portion of the circuit by summing one over 200 ohms with one over 1100 ohms. This gives us 0.005909 repeating inverse ohms, which is equal to one over the total resistance of this portion of the circuit. So to find the total resistance, we need to take one over this number. This gives us [169.23] ohms. So the total resistance of the parallel combination of resisters to the nearest ohm is 169 ohms.

What is the total resistance of the circuit to the nearest ohm?

Since we already found the total resistance of the parallel portion of the circuit in the previous part of the problem, we can draw an equivalent circuit to this one using that total resistance. The remaining resistors in the circuit are now in series. So we can find the total resistance by summing their resistances together. Once we add everything together, we’ll find that the total resistance of this circuit is 769 ohms.

What is the voltage across the 500-ohm resistor to two significant figures?

Before we address this part of the question, let’s first review how voltage and current behave in resistors in series and in parallel. This circuit has two resistors in series with a voltage source, 𝑣, and a current, 𝐼. The same current runs through the entire circuit, so the current across each resistor will be the same. And in this case, since there’s nothing else in the circuit, they will both be equal to the total current of the circuit. The voltage across each resistor will not be the same. According to Ohm’s law, which says that the voltage is equal to the current times the resistance, the voltage will be proportional to the resistance. Additionally, the voltage across each resistor in the circuit must sum to the total voltage of the circuit.

Now, we have a circuit with two resistors in parallel again with the voltage source, 𝑣, and a current, 𝐼. Now, the current in each resistor is not the same since the total current gets divided along each path in the circuit. Additionally, the current along each branch must sum to the total current of the circuit. The voltage across each branch of a circuit in parallel will be the same. Since there’s nothing else in the circuit besides the two resistors that are in parallel, in this case, the voltage across each resistor will be the same as the total voltage of the circuit. Now, let’s use what we know about circuits in parallel and in series. To find the voltage across the 500-ohm resistor.

Here, I’ve redrawn the portion of the circuit that contains the 500-ohm resistor. Since the 200-ohm resister is in parallel with the 600- and the 500-ohm resistor, the voltage across each branch must be the same. Here, I’m labeling the upper branch as a and the lower branch as b. We also know that the current of the upper branch and the lower branch, or 𝐼 a and 𝐼 b, must sum to the total current of the circuit, which is 100 milliamps. Also, 𝐼 b, the current across the lower path, will be equal to the current in the 500-ohm resister because the resistors in this path are in series. So if we can find 𝐼 b, we’ll have the current across the 500-ohm resister, which should help us find the voltage across the 500-ohm resistor.

We can use Ohm’s law and our knowledge that the voltage across the upper branch will be equal to the voltage across the lower branch. We can create an expression that relates the currents and resistances of both branches of this part of the circuit. Now that I’ve rearranged this expression, let’s plug in what we know. 𝑅 b, or the total resistance across the bottom path of the circuit, is 600 plus 500 or 1100 ohms. 𝑅 a, or the total resistance across the upper path of the circuit, is 200 ohms. This ratio is equal to 5.5. If we rearrange our expression relating 𝐼 a and 𝐼 b, we can solve for 𝐼 a giving us 100 milliamps minus 𝐼 b.

Let’s substitute this expression for 𝐼 a into our equation. We can now multiply the 𝐼 b on the bottom over to the other side and then add 𝐼 b to both sides giving us 6.5 𝐼 b is equal to 100 milliamps. So 𝐼 b is equal to 15.3846 milliamps. Now that we know 𝐼 b, we can finally find the voltage across the 500-ohm resistor.

We can solve for this voltage using Ohm’s law. The current in the 500-ohm resister is the current that we just solve for again because the 600- and the 500-ohm resistors are in series. This current is 15.3846 milliamps. The resistance of the 500-ohm resister is 500 Ohms. Before we can solve for the voltage, we need to convert the current from milliamps to amps, which we can do by dividing by 1000 because there’s 1000 milliamps in an amp. Now, we can solve for the voltage. This gives us 7.6923 volts. So if we round to two significant figures, we’ll see that the voltage across the 500-ohm resistor is 7.7 volts.

What is the current in the 600-ohm resistor to the nearest milliampere?

We’ve actually already solved for this current in the previous part of the problem. When we solve for 𝐼 b, we found the current in the 600-ohm resistor as well. Because the 500-ohm resistor and the 600-ohm resistor are in series, both of their currents will be equal to the current across that entire branch of the circuit. So rounding this number to the nearest milliampere, the current in the 600-ohm resistor is 15 milliamps.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.