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

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Discussion Overview

The discussion centers around calculating the power drawn from +9V and -9V batteries by an amplifier that draws 8mA from each power supply. Participants explore the interaction of currents and voltages in the context of amplifier power calculations, including the implications of using both positive and negative voltage supplies.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how the two currents (from +9V and -9V) interact, questioning whether they cancel, multiply, or sum up.
  • Another participant suggests that the circuit configuration is not clear and implies that assumptions about the circuit need to be made.
  • A third participant provides specific voltage and current values, attempting to apply the power formula P = VI but is uncertain about how to treat the currents from the two batteries.
  • One participant questions the representation of Vcc and Vee, noting the contradiction of having one positive and one negative voltage while being drawn in the same direction.
  • A later reply encourages focusing on the batteries themselves and suggests applying Kirchhoff's Current Law (KCL) to analyze the current flow, hinting at the relationship between the power delivered by the batteries and the power drawn by the amplifier.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to approach the calculation of power drawn from the batteries, with multiple competing views on the interaction of currents and the implications of the circuit configuration remaining unresolved.

Contextual Notes

There are limitations in the assumptions made about the circuit configuration and the interaction of the currents, which are not fully clarified in the discussion. The dependence on specific definitions of voltage and current in this context is also noted.

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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I take it we are supposed to guess what the circuit that you are talking about looks like
 
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?
 

Attachments

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Vcc and Vee are drawn as identical in direction yet you have said one is positive and one is negative. Why is that?
 
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.
 

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