[SCEN167 logo] Solution to 3/12 Quiz

In the circuit shown below, the 1.5 V battery is opposing the 6.0 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:
      6.0V - 1.5V = 4.5V, since the batteries are in series and oppositely directed.
    • Next find the effective resistance:
      50ohm + 25ohm = 75ohm, since the resistors are in series.
    • Finally, the current flowing can be found from Ohm's law:
      I = V/R = (4.5V)/(75ohm) = 0.060A
  2. What is the voltage across the 50 ohm resistor?
    • From Ohm's law:
      V = IR = (0.060A)(50ohm) = 3.0V
    • Additionally, the voltage across the 25 ohm resistor is:
      V = IR = (0.060A)(25ohm) = 1.5V
    • The sum of voltages is 4.5V as it must be -- answer checks.
  3. What is the power dissipated by the 25 ohm resistor?
    • Using P=I^2 R:
      P = (0.060A)^2 x 25ohm = 0.090W
    • Or using P=IV and the result from part 2:
      P = (0.060A)(1.5V) = 0.090W
    • Additionally the power dissipated by the 50 ohm resistor is
      P = (0.060A)(3.0V) = 0.18W, twice that of the 25 ohm resistor.
    • For a total power dissipation by the resistors of
      0.090W + 0.18W = 0.27W
  4. What is the power delivered by the 6.0 V battery?
    • Using P=IV for the battery:
      P = (0.060A)(6.0V)=0.36W
    • Where is the missing 0.09W?
      (Power provided by 6.0V 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 = (0.060A)(1.5V) = 0.090W, the difference!
[Quiz1 circuit]

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