6.1.2: Explorations

Exploration 1: Circuit Analysis

This Exploration begins with four identical lightbulbs connected to a battery (voltage is given in volts and current is given in amperes). Often, you will be asked to find the current through, the voltage across, the resistance, and/or the power consumed by a given bulb (or group of bulbs). To solve these types of problems you will use Ohm's law, \(V = IR\), and an equation for power, \(P=VI = I^{2} R= V^{2}/R\), where \(V\) is the voltage, \(I\) is the current, \(R\) is the resistance, and \(P\) is the power. You will also need to use two rules that are based on conservation laws:

1. current in = current out. Since charge is not created or destroyed (conservation of charge), charge flowing into some point must flow back out unless there is a circuit element that can store charge (a capacitor).
2. \(\Delta V\) through a complete loop \(= 0\). The electric force is a conservative force (which is why we can define an electrostatic potential: \(V\)). This means that if you start at one point in a circuit and add up all the potential increases and subtract all the decreases in potential as you trace out a loop, then when you get back to the place you started, the potential must be the same.

Let's use these rules for the initial circuit. The brightness of the bulbs is an indication of the current through the bulbs (brightness actually goes as \(I^{2}\)). Restart.

1. Rank the bulbs in order of brightness (and therefore in order of current through them), from highest to lowest.
2. Show the currents (in the data table) through the bulbs to check your answer. The arrows indicate the direction of the current through the circuit.
3. Find the current through bulb \(D\) by determining how much current must be coming into the node (the dot where the wires come together) above bulb \(D\). The current is coming from bulbs \(A\) and \(B\).
4. Check your answer by verifying that the current coming into the node below bulb \(D\) (from bulb \(C\) and bulb \(D\)) is equal to the current going out of the node and into the battery.

Now consider the voltage across various elements. Show the voltages (in the data table) across the bulbs.

5. Why is the voltage across bulb \(C\) the same as the voltage across the battery (think about tracing a path around the outside "loop" of the circuit)?
6. Why is the voltage across bulb \(A\) equal to the voltage across bulb \(B\)?
7. Find the voltage across \(D\) by picking a complete loop to trace around (battery \(\to\) bulb \(A\) \(\to\) bulb \(D\) \(\to\) battery) OR (battery \(\to\) bulb \(B\) \(\to\) bulb \(D\) \(\to\) battery) OR (bulb \(C\) \(\to\) bulb \(D\) \(\to\) bulb \(A\)) OR (bulb \(C\) \(\to\) bulb \(D\) \(\to\) bulb \(B\)) and finding the value of the voltage for bulb \(D\) that makes the change in potential equal to zero.
8. Using \(V = IR\), what is the resistance of bulb \(D\)? (Check that it is the same as bulbs \(A,\: B,\) and \(C\)).
9. What is the power dissipated by bulb \(D\)?

Exploration authored by Anne J. Cox.

Exploration 2: Lightbulbs

In this animation, you can close and open switches to determine the resistance of each lightbulb. The current and voltage readings show the current through the battery and the voltage across the battery (voltage is given in volts and current is given in amperes). In order to solve such problems, you can always use Kirchhoff loop equations. If it is possible, a faster way to work problems is in terms of the effective resistance of the network of resistors. You consider the resistors that are in parallel and resistors that are in series. It is, however, worth looking at the circuit in a bit more detail first, before diving into equations, to see if there is a way to understand the problem conceptually and to solve it faster. Restart.

Notice that when both switches are closed, bulbs \(A\) and \(B\) are dimmer than bulb \(C\). This should not be surprising because the current through bulb \(C\) is the sum of the currents through \(A\) and \(B\). Open one of the switches and leave the other one closed. Now bulb \(C\) is in series with one of the bulbs (which one?). Notice that the total current from the battery is less, but that either bulb \(A\) or bulb \(B\) is brighter than it was before.

1. Why?

Go back to the case in which both switches are closed and notice that bulbs \(A\) and \(B\) look to be the same brightness. If the brightness were exactly the same, they would have the same resistance.

2. With switch 1 open and switch 2 closed, what is the current?
3. What about with switch 1 closed and switch 2 open?
4. How does this "prove" that bulbs \(A\) and \(B\) are identical?
5. Furthermore, when one switch is open and one closed, how does the brightness of bulb \(C\) compare with the bulb it is in series with?
6. What does that indicate?

Now for some math.

7. Since \(R_{A} = R_{B}\), explain why, when both switches are closed, the effective resistance of the circuit is \(1/2R_{A} + R_{C}\). (Hint: with both switches closed, \(A\) and \(B\) are in parallel with each other).
8. When both switches are closed, use the voltage across the battery and the current through the battery to find the value of the effective resistance.
9. With one switch open and one switch closed, use the voltage across the battery and the current through the battery to find the value of the effective resistance. The effective resistance is equal to \(R_{A} + R_{C}\) (or \(R_{B} + R_{C}\)).
10. Solving these equations, you should find that all the bulbs are indeed identical (something you surmised from the brightness of the bulbs).

Notice that in this problem, trying to understand conceptually what was happening helped to guide the problem-solving process. Although the Kirchhoff loop rules will work, they are not necessarily the easiest way to solve a problem.

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

Exploration 3: Designing a Voltage Divider

Often with circuits, not only do you want to be able to figure out what a circuit that is already built is doing, you may want to design a circuit for a specific task. In this case our task is to design a circuit that is a voltage divider with a particular output voltage (voltage is given in volts and resistance is given in ohms). You have a \(12\text{-V}\) supply that can give you \(1\text{ W}\) of power, and you need a \(4\text{-V}\) output with as much power as possible. The resistors that you have can dissipate \(1\text{ W}\) of power. Restart.

To divide the voltage, we can put the power supply in series with two resistors and then use the voltage across one of the resistors to be our \(4\text{-V}\) output.

1. What ratio of resistors do you need to divide the supply voltage by one third? In other words, how many times bigger (or smaller) should resistor \(A\) be than resistor \(B\) to get an output of \(4\text{ V}\)? Try it.
2. Once the ratio is set up, do you have the maximum available power? To determine this, figure out the power used from the voltage source (\(P = V I\)). To get the maximum power (at a fixed voltage), should you increase or decrease the resistance in the circuit?
3. What is the limit on the total resistance (\(R_{A} + R_{B}\)) and, therefore, the limit on each resistor? Try it.
4. Try using a smaller value of resistance. Does the power supply burn up? (Fortunately, you can simply restart the animation and try again).
5. Double the values of \(R_{A}\) and \(R_{B}\). How much power does this circuit now draw from the battery?

Now that you have determined convenient values of \(R_{A}\) and \(R_{B}\) that produce a \(4\)-Volt output, replace the voltmeter with a lightbulb. (Adding a power-consuming circuit element is sometimes referred to as adding a "load.")

6. When this lightbulb is added, what is the voltage across the lightbulb?
7. Why is it less than \(4\text{ V}\)?
8. If you increase \(R_{A}\) and \(R_{B}\) more, what happens to the voltage across the lightbulb? Why? This is the reason voltage dividers like this are made from resistors that are as small as possible.

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

Exploration 4: Galvanometers and Ammeters

An ammeter measures current through a device and therefore must be in series with whatever element you want to find the current through. In this animation we will contrast the behavior of ideal and real ammeters by exploring the way a basic galvanometer works and finding how you can build an ammeter using a galvanometer (voltage is given in volts and current is given in amperes). Restart.

A galvanometer is a very sensitive meter that deflects when a small amount of current passes through it. (The current often goes through a coil that induces a magnetic field that causes an indicator needle to move. We use a red indicator bar instead of a needle.) In the galvanometer animation you can enter a source current and push the "galvanometer" button. The current source is shown using two interlocking circles, . The indicator bar on the right shows the maximum current that can pass through the galvanometer without damaging the instrument.

1. Change the source current so that the indicator is at \(50\%\). What is the current through the galvanometer and the voltage drop across the galvanometer? (This voltage drop is due to the internal resistance of the coil.)
2. What, then, is the maximum current that should go through the galvanometer (to just get the red bar to \(100\%\) of the screen)?
3. What happens if you exceed the maximum current rating?

We see that a galvanometer is a very sensitive current meter. They are often rated not by the maximum current, but by the internal resistance and the associated voltage drop at the maximum current.

4. Show that the internal resistance of this galvanometer is \(0.2\:\Omega\) and the voltage drop at maximum current is \(0.2\:\mu\text{V}\).

Suppose we want to use a galvanometer to measure currents up to \(1\text{ mA}\). We know that we want full-scale indication (bar at \(100\%\)) at \(1\text{ mA}\) and half-scale at \(0.5\text{ mA}\) and so we need for our meter to be made up of the galvanometer plus a resistor in parallel. This configuration is called an ammeter. For the galvanometer to just read full scale at \(1\text{ mA}\), only \(1\:\mu\text{A}\) of current can go through the galvanometer, and the other \(999\:\mu\text{A}\) must go through the parallel resistor.

5. If the voltage drop is \(0.2\:\mu\text{V}\), what value does the resistor in parallel need to have?
6. Try the value (for \(R_{x}\)) and then test by adjusting the power supply to see if you get the appropriate indication over the range of values (e.g., you should get half-scale indication for a source current of \(0.5\text{ mA}\); \(80\%\) of the bar would indicate a source current of \(0.8\text{ mA}\), etc.). Here use the "ammeter" button.
7. What would the ideal value of the internal resistance for an ammeter be and why?

Exploration authored by Anne J. Cox.

Exploration 5: Voltmeters

A voltmeter measures the voltage across a circuit element and therefore is put in parallel with that element. We can construct a voltmeter by placing a large resistor in series with the galvanometer indicated by  (an ammeter symbol) in the circuit because a galvanometer and an ammeter are essentially the same (see Exploration 30.4). In this example the galvanometer shows a full-scale indication at a current of \(1\:\mu\text{A}\), and the internal resistance of the galvanometer is \(0.2\:\Omega\).

1. What voltage across the galvanometer produces a full-scale reading?

If we want to measure battery voltages of up to \(2\text{ V}\), we'd want the galvanometer needle to give a full-scale indication at this voltage. This means that \(0.2\:\mu\text{V}\) must drop across the galvanometer (with a current of \(1\:\mu\text{A}\)), while \(1.9999998\text{ V}\) must drop across the series resistor. Restart.

2. Calculate the value of the series resistor required to produce a full-scale reading when the input voltage is \(2\text{ V}\). Test to see if you get the appropriate reading for a range of battery voltages. (Use the "set values" button.) Specifically, check that, for a battery voltage of \(1\text{ V}\), you get a half-scale reading on the indicator bar.
3. What would the ideal value of internal resistance for a voltmeter be and why?

Exploration authored by Anne J. Cox.

Exploration 6: RC Time Constant

In this animation you can close and open switches to see what happens to the voltage across the capacitor \(\color{red}{\text{(red)}}\), the voltage across the resistor \(\color{green}{\text{(green)}}\), and the total voltage across the capacitor plus resistor \(\color{blue}{\text{(blue)}}\). Initially, the capacitor is charged. After pushing "play, " you should throw the switches (voltage is given in volts and time is given in seconds). Restart.

Set the switches so that you can get a good graph of the capacitor discharging and charging.

1. How much time (approximately) does it take the capacitor to charge and discharge?
2. Double the battery voltage. How much time does it take to charge and discharge?
3. Double the capacitance and measure the time to charge and discharge.
4. Double the resistance and measure the time to charge and discharge.

The value of RC (resistance times capacitance) is the RC time constant for the circuit and is a characteristic time. Set the battery voltage to \(1\text{ V}\).

5. When the time equals \(R\ast C\) after throwing the switch, what is the capacitor voltage when it is discharging? When it is charging?
6. Compare your measurements to the values found from the equations for a charging or discharging capacitor:

\[\text{Charging: }V = V_{0} (1 - \text{e}^{-t/RC})\quad\text{Discharging: }V = V_{0}\text{e}^{-t/RC}\nonumber\]

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

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