Calculating Amplifier Power Draw from +9v and -9v Batteries - 8mA Current

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PhysicsThrow
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Homework Statement


Calculate the power drawn from the batteries by the amplifier. The batteries are +9v and -9v.
The amplifier draws a current of 8mA from each of its power supplies.

Homework Equations


P = VI

The Attempt at a Solution


I'm just a little stumped as to what to do for the current. I know V = 18v but I'm not sure how the 2 currents interact with each other. Do they cancel out? Do they multiply or sum up?
I'm sure I can finish the rest if I understand that.
 

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cj1Bzpv.png


VCC = 9V
VEE = -9V
VO = 6v rms
VI = 2v rms
RL = 900Ω
Amplifier draws 8mA from each battery.
Input sinusoidal current of 1v rms.

So the terms relevant to the amplifiers power draw are
VCC, VEE and the 8mA current from each battery.

I assume I just calculate using P = VI = (9*2)*I
But I can't work out what I should equal. Is it (8mA*2) or (8mA+-8mA=0)? Or is it just 8mA?
 

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PhysicsThrow..

Hint: Forget the amplifier, just look at the batteries. What current is flowing through each battery? If you can't see it, move your symbols for ICC and IEE nearer the batteries on the drawing. How much power are the batteries delivering? That's the same as the amplifier is drawing (with no signal)

Still stuck? Apply KCL to he node between the two batteries. Is there any current flowing in the ground wire?
 
Last edited:
PhysicsThrow said:

Homework Statement


Calculate the power drawn from the batteries by the amplifier. The batteries are +9v and -9v.
The amplifier draws a current of 8mA from each of its power supplies.

Homework Equations


P = VI

The Attempt at a Solution


I'm just a little stumped as to what to do for the current. I know V = 18v but I'm not sure how the 2 currents interact with each other. Do they cancel out? Do they multiply or sum up?
I'm sure I can finish the rest if I understand that.

The batteries are the power supplies I assume.

So the answer is too obvuious to give any hints.