April 30 RC Circuits

Introduction: Today we will explore and analyze RC circuits, and also charge buildup and decay in capacitors. An RC circuit is also known as a resistor capacitor circuit and is an electric circuit composed of resistors and capacitors driven by a voltage or current source. 

We began class with our first activity, in which we were given a capacitor, batteries, bulbs and a switch. (pictured below)

Activity:  Capacitors, Batteries, and Bulbs

a. Connect a rounded #14 bulb in series with the 0.47 F capacitor, a switch, and the 4.5 V power supply.  Describe what happens when you close the switch.  Draw a circuit diagram of your setup.

b. Now, can you make the bulb light again without the battery in the circuit?  Mess around and see what happens.  Describe your observations and draw a circuit diagram showing the setup when the bulb lights without a battery.

What we found for part a is that the light bulb lit up and for part b the light bulb lit then dimmed. For part a the light bulb lit because there was voltage from the battery flowing through the circuit once the switch was closed. In part b, since there was a capacitor in the circuit, the capacitor had a charge buildup. The charge came from the battery when it was connected to the circuit. When the battery was taken out, this charge buildup gave electric energy to the light bulb causing it to stay lit for a short time without a battery being connected, however the light bulb eventually dimmed and went out because the charge buildup was exhausted.

c. Draw a sketch of the approximate brightness of the bulb as a function of time when it is placed across a charged capacitor without the battery present.  Let t = 0 when the bulb is first placed in the circuit with the charged capacitor.  Note:  Another way to examine the change in current is to wire an ammeter in series with the bulb.

Our graph for part c and our summary for parts a and b are in the picture below.

e. What happens when more capacitance is put in the circuit?  

We said that an increase in capacitance would take a longer time to charge than if there was less capacitance.


For our next experiment, we were asked to create a graph that would plot the capacitor's voltage over time. Using logger pro, a capacitor, a resistor, battery and voltage probe, we created a circuit and got to work. We began by hooking up the resistor to the capacitor, and the voltage probe to logger pro; the circuit was connected in series. We zeroed out the voltage,   then connected to the battery in order to create our graph. The following six pictures are our results from our experiment.

In summary of the experiment from the experiment of above: we found that our graph was not linear, but instead it was exponential. By finding the equation of the line and mathematical reasoning based on the application of Ohm's law as well as the definition of current and capacitance, we found our  equation for exponential decay (pictured below).

The Theoretical RC Decay Curve

Using Ohm's Law V = IR, the definition Capacitance C= Q/V, and the definition of current I = dQ/dt, we can rewrite this last equation so that the derivative of the charge is proportional to the negative of the amount of charge on the capacitor. Pictured below is our work for this derivation.

Our solution for our derivation should have looked liked this (picture above).

Professor Mason then gave us two graphs, (not pictured), one was a voltage vs. time graph and the other was a brightness vs. time graph. He then asked us to predict what a current vs. time graph would look like based on the two other graphs he gave us. Our prediction is pictured below.

We found that power is directly proportional to brightness and power is I*V. From this we can conclude that as power decreases and voltage increase, current decreases over time.

RC Circuit Problem

A 100 mF capacitor is connected in series with a 10.0 Ω resistor. This combination is connected in parallel with a 25.0 Ω resistor. Both branches are then connected in parallel to a 4.50 V battery that can be switched on and off. The capacitor starts off fully discharged. 

 Draw a picture of the circuit

Loop 1 is just that of a simple RC circuit when the capacitor is being charged up and has a time constant τ₁ = R ₁ C . Thus we can use the generic equations associated with charging a capacitor. The basic ones involve the current through the resistor and the charge on the capacitor. 

The answer for loop one is 1.

Loop 2 is just that of a simple DC circuit with resistor R₂. Loop 2 does not have a time constant associated with capacitance in the loop. The will reach is maximum value almost instantaneously and remain constant until the switch is opened.

Our answer for loop two is 0.

Here I wrote down the definition of voltage for decay curve when its charging and discharging.

For our next activity we were asked the following questions: 

A) What is the maximum charge that the capacitor can attain after the switch is closed?

B) When will the voltage drop across the 10.0 Ω resistor be equal to 1.50 V after the switch is closed? 

a) We found Q by using our definition of capacitance. We found Q to be .45C.

b) by applying the theory of the RC decay curve we found the time it would take for the voltage to drop across the 10 ohm resistor to be equal to 1.50 V after the switch is closed, which turned out to be 1.1 seconds.

Final question:

C) If the switched is opened, what will be the value of the new time constant and in which direction will the current flow through the 10.0 Ω resistor? 

We found tao to be 3.5 and its flowing in the clockwise direction.