6.1.1: Illustrations

Illustration 1: Complete Circuits

This Illustration examines open and closed circuits by considering a circuit composed of three lightbulbs. These lightbulbs behave very much like resistors and are often represented using the resistance symbol, . Restart.

Note that even though these lightbulbs are identical, their brightness can vary. The relationship between voltage, current, and brightness is discussed later in this chapter, but you should notice that the brightness of a bulb depends on the voltage across it.

A burned-out filament results in an open circuit in that branch and is equivalent to a very large resistance. First predict what will happen when one or more lightbulbs burn out and then click on the links to test your prediction. The animation will show you both the bulbs (which ones light and which ones do not) and the voltages (voltage is given in volts).

Remember that if there is a voltage drop across a "good" lightbulb (one that is not broken), it should light up. Notice that when bulb 1 is out, none of the lightbulbs light. If only bulb 2 or bulb 3 is out, notice that the other bulbs light. If bulbs 2 and 3 are both out, notice that bulb 1 does not light up. Make sure that you can explain these observations and the associated voltage readings.

By looking at the voltage across each bulb (resistor), you should be able to analyze the circuit. Notice that you cannot determine which lightbulb is broken simply by looking at the bulbs because multiple conditions produce the same visual clues.

Illustration authored by Anne J. Cox.

Illustration 2: Switches, Voltages, and Complete Circuits

This Illustration allows you to control the current flow through a circuit using switches (voltage is given in volts). Restart.

You can open and close switches to turn the lightbulbs on and off. (How is this similar to a bulb breaking in Illustration 1?) Notice that when bulb 1 is dark (\(S1\) open and the other switches closed), there is no complete path for the current, so none of the bulbs light. Furthermore, there must be \(0\text{ V}\) across bulbs 2 and 3, and there is \(10\text{ V}\) across the switch \(S1\). Similarly, when bulb 2 is dark (\(S2\) open and the other switches closed), there is a complete path for current flow through bulbs 1 and 3 (and the voltage across each is \(5\text{ V}\)). The same reasoning applies to bulb 3 when switch 3 is open. (When all the switches are open, the voltmeter readings can have any value, as long as the sum adds up to \(10\text{ V}\).)

Illustration authored by Anne J. Cox.

Illustration 3: Current and Voltage Dividers

This Illustration shows two different configurations of resistors connected to a battery (voltage is given in volts, current is given in amperes, and resistance is given in ohms). Restart.

Start with the voltage divider animation. The circuit shows an ideal battery supplying current to a \(100-\Omega\) resistor in series with a variable resistor, \(R_{A}\). When the resistance of \(R_{A}\) is equal to \(100\:\Omega\) (the value of the top resistor), the voltage is equally divided across the two resistors. As you increase \(R_{A}\), what happens? What happens as you decrease \(R_{A}\)? The current from the battery also changes in this process; however, notice that the current through the top resistor and \(R_{A}\) are always equal. This is because the current through the top resistor must also go through \(R_{A}\).

Now try the current divider animation. The \(100-\Omega\) resistor is now in parallel with \(R_{A}\). When the resistance of \(R_{A}\) is equal to \(100\:\Omega\) (the value of the fixed resistor), the current is equally divided between the two branches of the circuit. As you increase \(R_{A}\), what happens? What happens as you decrease \(R_{A}\)? The current from the battery also changes in this process, but the voltage across the two resistors is the same because we have assumed that the battery is capable of supplying a large amount of current and because the wires are assumed to have negligible resistance. If you added a third resistor in parallel with the other two, the current from the battery would increase since the battery needs to supply current to that resistor (with the same voltage drop) as well.

Many students think that if you add additional resistors to circuits such as the ones just seen, the current from the battery must decrease (for the voltage to stay the same). Notice, however, that resistors added in parallel increase the total current from the battery, while resistors added in series reduce the total current from the battery.

Illustration authored by Anne J. Cox.

Illustration 4: Batteries and Switches

In this Illustration you can close and open switches to see what happens to the two identical bulbs. The table gives the voltage across the lightbulbs and the brightness of the bulbs indicates the relative current flow (voltage is given in volts). The batteries are all identical. Restart.

Notice that \(S1\) and \(S2\) cannot be closed or opened at the same time. What would happen if both \(S1\) and \(S2\) were open? What if both \(S1\) and \(S2\) were closed? (Which situation is bad for battery \(B1\), and why?) If \(S1\) and \(S2\) are both open, no current flows, but if \(S1\) and \(S2\) are both closed, this creates a short circuit. There is essentially no resistance in the path connecting the two terminals of the battery, so as much current as possible flows from the battery (quickly discharging it).

As you switch \(S1\) and \(S2\) together (one open and one closed), what happens to the lightbulbs? Why? (That is, what is the difference between the two circuits?) Notice that when you put another battery in series, the potential difference (voltage) across each bulb increases, as does the current (and the bulbs are brighter).

With \(S1\) closed and \(S2\) open, now close \(S3\). What happens? What is the voltage across each bulb? Why doesn't it change? You have added a battery to the circuit, but nothing happened. Why? Now close \(S1\) and open \(S2\). What happens? Why?

Now look at the voltage across bulb \(B\). When \(S1\) is closed, \(S2\) is open, and \(S3\) is open, the voltage across bulb \(B\) is \(9\text{ V}\). When you close \(S3\), the voltage across bulb \(B\) is still \(9\text{ V}\). With \(S3\) closed, the voltage across bulb \(B\) is always \(9\text{ V}\), no matter what is happening with \(S1\) and \(S2\). When you put batteries in parallel, you don't increase the total voltage. In fact, you need to be careful of adding batteries in parallel because if you have two batteries in parallel with each other at different voltages, you will end up with a great deal of current. This is why you have to be careful using jumper cables: You put two batteries in parallel with each other and even the slightest difference in voltage means a lot of current flows (because the jumper cables don't have much resistance. The current is limited by the internal resistance of the batteries). Of course this also works to your advantage to quickly recharge the dead battery.

Illustration authored by Anne J. Cox.

Illustration 5: Ohm's "Law"

Ohm's law is not a "law" in the sense of "laws of physics" (for example, Newton's laws, conservation of energy, conservation of charge, etc.). Instead, it describes a linear relationship between current and voltage that holds true for some circuit elements, namely resistors. There are, however, other circuit elements that do not follow Ohm's law. Vary the voltage of the source and note the linear current-voltage relationship for the resistor. Compare this to the nonlinear response of a diode and a lightbulb (voltage is given in volts and current is given in amperes). Restart.

• Resistor
• Diode (voltage scale is tenth of Volts)
• Real lightbulb

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J Cox.

Illustration 6: RC Circuit

In the animation you can close and open switches to see what happens to the lightbulb. Restart.

When the animation begins, the capacitor is initially charged. Push the "play" button and then open/close the switches. Watch what happens to the lightbulb. When the lightbulb goes out (is dark), throw the switches again. What happens? You should observe that the lightbulb is bright initially and always eventually goes out (even when there is a battery in the circuit). Notice, too, that the lightbulb is the brightest right after you throw the switches. This means that the current is the biggest. When the bulb is out, however, the current is zero. From these observations, what then is the voltage across the bulb as a function of time after the switch is thrown?

Show the graph of voltage vs. time. The voltage across the lightbulb \(\color{green}{\text{(green)}}\), the voltage across the capacitor \(\color{red}{\text{(red)}}\) and the total voltage across both \(\color{blue}{\text{(blue)}}\) are shown. What does the graph look like for the situation when the capacitor is charging (capacitor and resistor in series with the battery)? Notice that the lightbulb voltage plus the capacitor voltage equal the total voltage and that current flows (the lightbulb lights) until the capacitor voltage equals the battery voltage. What does the graph look like when the capacitor is discharging (battery not in the circuit with the capacitor and the resistor)? Notice that the capacitor and resistor voltages are equal and opposite so that the sum of their voltages is zero. A negative voltage across the bulb simply means that current is flowing in the other direction as the capacitor discharges down to \(0\text{ V}\). So as the capacitor charges, the current flows from the battery through the resistor and charge builds up on the capacitor, but as the capacitor discharges, the current flows from the capacitor through the resistor until the capacitor has no charge on its plates.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian.

Illustration 7: The Loop Rule

Kirchhoff's loop rule states that the sum of all the potential differences around a closed loop equals zero. In other words, \(\sum\Delta V=0\) for a complete loop. Restart.

In the circuit shown, current from the battery flows through the resistors before returning to the battery. This illustration follows a hypothetical charge as it flows through the upper of the parallel resistors. This is, of course, just a simulation. Current also flows through the lower resistor, and the current through these two resistors is not the same. In fact, an accurate microscopic simulation would need \(~10^{20}\) electrons moving counterclockwise around the circuit. Because an electron-flow representation of current would be awkward, we use the standard definition of current and show a hypothetical unit of positive charge flowing out of the positive battery terminal, through the resistors, and into the negative terminal.

In a circuit, there are charges moving through the potential differences from the battery and across the resistor. So another way to state the loop rule is that, when a charge goes around a complete loop and returns to its starting point, its potential energy must be the same. Positive charges gain energy when they go through batteries from the \(-\) terminal to the \(+\) terminal, and they give up that energy to resistors as they pass through them.

Use the loop rule to determine the current through the battery in a circuit consisting of a \(16\)-volt battery connected to a set of three resistors: a \(2-\Omega\) resistor in series with a parallel combination of a \(2-\Omega\) resistor and a \(3-\Omega\) resistor.

Now consider a Kirchhoff loop consisting of the battery and the two \(2-\Omega\) resistors. It doesn't matter where we start, as long as we come back to the same spot. Let's go clockwise around the loop starting at the bottom left corner.

\[+16\text{ V} - (2\:\Omega )\ast I - (2\:\Omega )\ast 3I/5 = 0\nonumber\]

\[+16\text{ V} = (10\:\Omega )\ast I/5 + (6\:\Omega )\ast I/5\nonumber\]

\[+16\text{ V} = (16\:\Omega )\ast I/5\nonumber\]

This gives \(I = 5\text{ A}\).

Run the animation and follow the energy of the unit charge as it passes through each circuit element. Each voltage drop represents the amount of energy that is lost or gained when the charge passes through a circuit element. This demonstrates that, as charge flows around a complete loop, the gains in energy are always offset by the losses. The total change in energy is zero.

Illustration authored by Andrew Duffy.
Script authored by Andrew Duffy.

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