Practice Problems: Short Circuits Solutions

1. (moderate) Given standard lab equipment, how would you determine the emf and the internal resistance of a standard flashlight battery? Write a procedure that could be followed by a fellow student.
The following procedure is one method that can be followed:
Step 1: Connect a circuit that contains the battery, an ammeter, and a resistor of known resistance (R₁). Choose a resistor that will allow for a reasonable current.  
Step 2: Use the ammeter to determine the current (I₁). Record this value.
Step 3: Use a loop rule to relate the parameters:
ε - I₁r - I₁R₁ = 0
Step 4: Replace the known resistor with a different known resistor (R₂). Repeat steps 1 and 2. Record the results.
Step 5: Create a second loop rule to relate the parameters again:
ε - I₂r - I₂R₂ = 0
Each of the loop rules contain the same unknowns, ε and r. Use the two equations to solve for the unknowns.

2. (moderate) A single loop circuit consists of a battery (10 v with 1 Ω of internal resistance) and a load resistor (100 Ω). A single piece of wire is also available that can short circuit the battery. Determine the terminal voltage of the battery before and after the short is applied.
Before the short:
Total R = 101 Ω
Find the current:
V = IR
10 = I(101)
I = 0.099 A
Terminal Voltage = emf - Ir = 10 - (0.099)(1) = 9.9 v
After the short:
V = Ir (for loop of emf and internal resistor)
10 = I(1)
I = 10
Terminal voltage = emf - Ir = 10 - 10(1) = 0

3. (easy) A flashlight battery (emf of 1.5 v) is shorted with a wire of almost zero resistance. The current in the wire is 28 A. Find the internal resistance of the battery.
Since the internal r is the only available resistance:
emf = Ir
1.5 = (28)r
r = 0.054 Ω

4. (moderate) A 2.0 Ω resistor is attached across a 12 v battery. The terminal voltage is measured to be 8.5 v. What is the internal resistance of the battery?
emf = I(R + r)
12 = I(2 + r)
I = 12/(2 + r)
Terminal voltage = emf - Ir
8.5 = 12 - (12/(2 + r))r
-3.5 = -12r/(2 + r)
-3.5(2 + r) = -12r
-7 - 3.5r = -8.5r
-7 = -8.5r
r = 0.82 Ω

5. (moderate) A simple circuit consists of a 12 v battery (whose internal resistance is 0.1 Ω) and a 100 Ω resistor. Compare the power dissipated by the battery before and after it is short-circuited.
Prior to the short circuit:
Total resistance = R_T = 100.1 Ω
ε = IR_T
12 = I(100.1)
I = 0.12 A
(This is the current through the battery)
P = I²r = (0.12)²(0.1) = 0.00144 watts dissipated by the battery
After the short circuit:
ε = Ir
12 = I(0.1)
I = 120 A
(This is the new current through the battery)
P = I²r = (120)²(0.1) = 1440 watts dissipated by the battery
This short circuit will cause severe heating of the battery and can be dangerous.

6. (hard) The sketch below shows two batteries (with internal resistance) connected to a load resistor. If the internal resistance of the lower battery is 0.016Ω and that of the upper battery is 0.012Ω, determine the resistance (R) if one of the batteries was found to have no terminal voltage. Also determine which battery can show this condition.


The total resistance in the circuit is R_T= R + 0.028Ω
If the terminal voltage across the lower battery is zero:
12 - Ir₁ = 0
12 - I(0.016) = 0
I = 750 A
If this is the case, the entire circuit loop would show:
24 = IR_T = 750(R + 0.028) R = 0.004Ω
If the terminal voltage across the upper battery is zero:
12 - Ir₂ = 0
12 - I(0.012) = 0
I = 1000 A
If this is the case, the entire circuit loop would show:
24 = IR_T = 1000(R + 0.028)
There is no solution for R in this case.
Thus R = 0.004 Ω and the lower battery is the one with no terminal voltage.

7. (hard) A battery (r = 0.5Ω) is connected to x number of lightbulbs (each the same with R = 15 Ω). The bulbs are all connected in parallel with the battery. The terminal voltage is measured to be ½ the emf. Find x.
R_bulbs = 1/(x/15) = 15/x
R_eff = R_bulbs + r
R_eff = (15/x) + (0.5)
Find the emf
emf = I(R_eff)
emf = I[(15/x) + 0.5]
Now find x:
Terminal voltage = ½emf = emf - Ir
½(I[(15/x) + 0.5] = I[(15/x) + 0.5] - Ir
15/x + 0.5 = 30/x + 1 - 2(0.5)
0.5 = (30/x) - (15/x) = 15/x
x = 15/0.5 = 30
Another way of thinking about this is that the load resistor and the internal resistance must have the same value to that each has the same voltage drop. This would allow the terminal voltage to be half the emf. 30 bulbs with 15 Ω of resistance (all in parallel) have an effective resistance of 0.5 Ω, just like the internal resistance.