Black Box Power Supply
In this experiment, you will determine the internal resistance, Rint of a power supply (the “black box”) as well as
the maximum power output for the power supply.
Background
Power supplies and batteries have three important operating parameters: terminal voltage Vt, emf, ε, and internal
resistance, Rint. These parameters are important when selecting a battery or power supply for maximum
efficiency. If these parameters are unknown then we can call the power supply a “black box”.
A battery can be though of as a source of ε connected in series with an internal resistor with resistance Rint.
When a load is placed across the battery’s terminals, current flows through the internal resistor which causes a
voltage drop across the terminals. More current means a greater voltage drop. We can use Ohm’s law to relate
Vt, ε, and Rint.
Vt = ε − IRint
The power dissipated by a resistor is defined as the rate at which charges lose electric potential energy (qΔV) in
going through the resistor.
P=
dU d
dQ
= (qΔV ) =
ΔV = IΔV
dt
dt
dt
We can use Ohm’s law to come up with alternative ways of calculating the power:
(ΔV ) 2
P = IΔV = I R =
R
2
Complete the following derivation and include it in the INTRODUCTION section of your lab report.
Step 1: Draw a circuit diagram showing a battery with emf ε, its internal resistance Rint (you can consider the
internal resistance as a resistor connected in series with the battery), and a load resistor with resistance Rload.
Step 2: Use Kirchoff’s Loop Rule to solve for the current in the circuit in terms of ε, Rint, and Rload. Call this
“Equation 1”.
Step 3: Using the fact that the power dissipated by a resistor can be written as P = I2R, come up with an
equation for the power dissipated by the load resistor in terms of ε, Rint, and Rload. Call this “Equation 2”.
Step 4: By taking the derivative of Equation 2 with respect to the load resistance, show that the maximum
power is delivered to the load resistance, when the load resistance is equal to the internal resistance of the
battery.
Experiment
1. Connect the black (or blue) box power supply in series with the variable resistor and a multimeter set to
measure current. Then connect another multimeter set to measure voltage in parallel with the variable resistor.
2. By adjusting the slider on the variable resistor, you are changing the load resistance Rload. The current and
voltage will also change. Adjust the slider until you get the smallest possible current. Now, take as many
measurements of voltage and current as you can, spacing each measurement by 0.25 mA. For instance, if your
smallest current is 4.0 mA, you would take measurements for 4.0 mA, 4.25 mA, 4.5 mA … all the way up to
your largest current. Record the current, voltage, Rload (calculated from Ohm’s law), and the power dissipated
by the load resistor in a table.
3. Make a plot of power vs. Rload, and use your plot to determine the internal resistance of the black box power
supply.
4. The black boxes have an internal resistance of 82-Ω and the blue boxes have an internal resistance of 100-Ω.
Compare your value for internal resistance with the true value, and calculate a percent error.
Questions
1. Well designed power supplies are “regulated”, meaning that they maintain a constant output voltage
regardless of the resistance of the load. In other words, ε and Vt are the same and remain equal for any current
drain. What would be the effective internal resistance of such a power supply? Explain.
2. On a warm day a car battery reads 13.0 volts when its current is zero. Find the terminal voltage Vt when the
starter draws 100 Amperes, assuming the internal resistance of the battery is 0.025-Ω.
3. When the battery is cold, the internal resistance increases to 0.10-Ω. What is the terminal voltage Vt when
the starter draws 100 Amperes? Will the car start? (a starter requires approximately 9-10 V).