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!
[Quiz1 circuit]

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Last updated Oct. 10, 1996.
Copyright George Watson, Univ. of Delaware, 1996.