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.
- 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
- 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.
- 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
- 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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