Solution to 10/8 Quiz

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!