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!

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