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

In summary, to calculate the power drawn from the batteries by the amplifier, we can use the equation P = VI and consider the current flowing through each battery, which is 8mA. This results in a power of 144mW being drawn from each battery. This assumes that the batteries are connected in series, with the amplifier drawing 8mA from each one.
  • #1
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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  • #2
I take it we are supposed to guess what the circuit that you are talking about looks like
 
  • #3
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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  • #4
Vcc and Vee are drawn as identical in direction yet you have said one is positive and one is negative. Why is that?
 
  • #5
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?
 
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  • #6
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.
 

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

What is amplifier supply current?

Amplifier supply current refers to the amount of electrical current required by an amplifier to function properly. It is an important specification for amplifiers as it determines the amount of power they can deliver to the output signal.

Why is amplifier supply current important?

Amplifier supply current is important because it affects the performance and capabilities of the amplifier. The higher the supply current, the more power the amplifier can deliver to the output signal, resulting in better sound quality and louder volume.

How is amplifier supply current measured?

Amplifier supply current is measured in amperes (A) or milliamperes (mA) and is typically listed in the amplifier's specifications. It can also be calculated by dividing the power supply voltage by the load resistance.

What factors affect the amplifier supply current?

The amplifier supply current can be affected by various factors such as the type of amplifier circuit, the load placed on the amplifier, and the power supply voltage. Higher power supply voltages and lower load resistance can result in higher supply current.

How does amplifier supply current impact power consumption?

Amplifier supply current directly affects power consumption as it determines the amount of power the amplifier is drawing from the power source. Higher supply current means higher power consumption, which can result in increased energy costs and potential strain on the power supply.

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