# Solution to 10/8 Quiz

In the circuit shown below, the 1.5 V battery is opposing the 9.0 V battery as they are positioned. The total voltage of the two batteries will therefore be found by subtracting.
1. What is the current flowing in the circuit?
• First find the effective battery:
9.0V - 1.5V = 7.5V, since the batteries are in series and oppositely directed.
• Next find the effective resistance:
50ohm + 100ohm = 150ohm, since the resistors are in series.
• Finally, the current flowing can be found from Ohm's law:
I = V/R = (7.5V)/(150ohm) = 0.050A
2. What is the voltage across the 50 ohm resistor?
• From Ohm's law:
V = IR = (0.050A)(50ohm) = 2.5V
• Additionally, the voltage across the 100 ohm resistor is:
V = IR = (0.050A)(100ohm) = 5.0V
• The sum of voltages is 7.5V as it must be -- answer checks.
3. What is the power dissipated by the 100 ohm resistor?
• Using P=I^2 R:
P = (0.050A)^2 x 100ohm = 0.25W
• Or using P=IV and the result from part 2:
P = (0.050A)(5.0V) = 0.250W
• Additionally the power dissipated by the 50 ohm resistor is
P = (0.050A)(2.5V) = 0.125W, half that of the 100 ohm resistor.
• For a total power dissipation by the resistors of
0.250W + 0.125W = 0.375W
4. What is the power delivered by the 9.0 V battery?
• Using P=IV for the battery:
P = (0.050A)(9.0V)=0.450W
• Where is the missing 0.075W?
(Power provided by 9.0V battery) - (total power dissipated by resistors)
• Current is being pushed the "opposite" way through the 1.5V battery
-- it is being "charged" (absorbing power from the circuit)
-- rather than discharging (providing power to the circuit).
P = IV = (0.050A)(1.5V) = 0.075W, the difference!

Top of page

Comments, suggestions, or requests to ghw@udel.edu.

"http://www.physics.udel.edu/~watson/scen103/qoln1.html"
Last updated Oct. 10, 1996.
Copyright George Watson, Univ. of Delaware, 1996.