- 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**

- First find the effective battery:
- 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**.

- From Ohm's law:
- 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

- Using P=I^2 R:
- 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!

- Using P=IV for the battery:

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Last updated Oct. 10, 1996.

Copyright George Watson, Univ. of Delaware, 1996.