- 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**

- First find the effective battery:
- 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**.

- From Ohm's law:
- 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

- Using P=I^2 R:
- 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!

- Using P=IV for the battery:

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