Values of Currents & Potential Differences in an Electric Circuit

Instructor: Matthew Bergstresser
In this lesson, we will design an electrical circuit including a battery, some resistors, a few switches and a capacitor. Then, we will change the circuit by opening and closing various switches and make predictions of voltages and currents.

Making Changes

Imagine baking chocolate chip cookies from scratch using a trusted recipe. You can predict before those cookies come out of the oven what they will taste like. Now imagine deciding to change the recipe a bit; maybe you want to reduce the sugar amount. You can predict how the cookies might taste with the change and evaluate your prediction when you actually eat them. We are going to do the same thing except with electrical circuits. We will design a circuit and then change it by adjusting switch positions and then evaluate our predictions of what will change in terms of current and voltage through a resistor that receives current in both switch arrangements.

RC Circuit

An RC circuit is one that includes resistors and capacitors. A resistor resists the flow of electric current, and a capacitor stores electric charge. Let's start analyzing an RC circuit that contains one battery, three switches, four resistors, and one capacitor.

Circuit with initial switch configurations

Arrangement 1

The solid and dotted lines show how the switches can be closed and opened. A solid line is a closed switch, and a dotted line is an open switch. Initially, switch 1 is open and the capacitor is uncharged. Switch 2 is in the left position, and Switch 3 is closed. Let's close Switch 1, open Switch 3, and keep Switch 2 in the left position. We will determine the initial instantaneous current and voltage across the 100 Ω and 200 Ω resistors. We'll call this switch configuration Arrangement 1.

Arrangement 1. The blue arrows show the direction of current flow.

We will use Kirchhoff's loop rule to determine the currents through the resistors. Kirchhoff's loop rule states that the sum of voltages in a closed loop must equal zero. We will start at the bottom left corner of the circuit and move counterclockwise following the blue arrows, which is I1.

-200I1 -100I1 + 60 = 0

-300I1 + 60 = 0

300I1 = 60

I1 = 0.2 amps

This is the current through the 200 Ω resistor.

The voltages across the 200 Ω resistor can be calculated with Ohm's law, which is V = IR. Let's calculate this voltage.

V100 = (0.2 amp)(100 Ω) = 20 V

We can collect data from the circuit using a multi-meter to verify our theoretical results. When we connect the multimeter across the 20 Ω and 50 Ω resistors, and the 4 µF capacitor, the voltage reads zero. This makes sense because there is no current through them! When we connect the multimeter across the 200 Ω resistor we get exactly what we calculated!

Arrangement 2

Now we will move Switch 2 to the right position, and close Switch 3. We will predict that this switch configuration will allow current through the capacitor and the 50 Ω resistor. Since the current from the battery splits, we should get a smaller current through the 200 Ω resistor than we got for the current through it in switch Arrangement 1. We will calculate the currents the instant these switch positions are changed. Arrangement 2 shows the new currents.

Arrangement 2

Notice we have three currents now. We have to use Kirchhoff's junction rule along with two loop equations to determine these currents. Kirchhoff's junction rule states that current into a junction has to equal current out of a junction. Let's start with a junction equation for the junction just to the right of Switch 1.

I2 = I3 + I4

Now we will write the equation for loop 1 (L1), which is counterclockwise around the outside of the circuit.

-20I4 - 50I2 + 60 = 0

Next is loop 2 (L2).

200I3 - 20I4 + = 0

We can now use these equations to solve for the three currents, which are shown in Table 1.

Table 1
I2 I3 I4
0.88 amp 0.08 amp 0.8 amp

When we connect the multimeter across these resistors we get the exact values we predicted. I3 is the current through the 200 Ω resistor. That resistor is getting a 60% reduction in current compared to the current through the same resistor as in switch Arrangement 1. Our prediction was correct!

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