# Help Understanding Power in Circuits

• HydroGuy
In summary, the amount of power drawn from a voltage source is determined by the internal resistance of the source and the impedance of the load. To design for low power, a high load impedance is desirable. This is because according to Ohm's Law, the current drawn from a power source is inversely proportional to the load impedance. Additionally, minimizing bias currents in active circuits can also help reduce overall current consumption.
HydroGuy
In University we almost always deal with voltage sources that can supply whatever current needed. What determines the amount of power "drawn", is it purely the resistance? How do we design for low-power?

The current drawn from a power source depends on the internal resistance of the power source and the impedance of the load.

So, you don't have to design for low power. If you put a 1000 ohm resistor across a 12 V car battery, it will deliver 12 mA even though the battery can deliver hundereds of amps.
(This is Ohm's Law... I = V / R = 12 /1000 = 12 mA.)

The internal resistance of the power source decides the maximum current the supply can deliver. If this is not important, then it is convenient to regard the supply as perfect and assume the voltage will remain stable with varying load currents.

So... the key to low power is to have a very high load impedance? I don't think I'm comprehending this correctly.

HydroGuy said:
So... the key to low power is to have a very high load impedance?

Yes, per Ohm's law.

And for active circuits, you try to minimize the bias currents, in order to minimize overall current consumption. That brings on a host of issues, however.

Here's an intro article with lots of links to other references:

http://en.wikipedia.org/wiki/Low-power_electronics

.

So... the key to low power is to have a very high load impedance?

Yes, exactly. The power actually drawn from a power source doesn't have to be as much as the source is capable of.
A car battery is quite capable of delivering only 1 mA if the load is 12000 ohms. This is only 12 milliwatts!
Or, it could deliver 50 amps if the load was 0.24 ohms. This would be 600 watts. (12 volts * 50 amps = 600 watts).

## 1. What is power in a circuit?

Power in a circuit refers to the rate at which energy is transferred or converted in an electrical system. It is measured in watts (W) and is calculated by multiplying the voltage by the current.

## 2. How do I calculate power in a circuit?

To calculate power in a circuit, you can use the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. Alternatively, you can also use the formula P = I^2 * R, where R is the resistance in ohms.

## 3. What is the relationship between power, voltage, and current in a circuit?

The relationship between power, voltage, and current in a circuit is described by Ohm's Law, which states that P = VI. This means that power is directly proportional to both voltage and current. In other words, increasing the voltage or current in a circuit will result in an increase in power.

## 4. How does power affect the components in a circuit?

The amount of power in a circuit can affect the components in different ways. For example, too much power can cause components to overheat and potentially fail, while too little power may not provide enough energy for the components to function properly. It is important to properly calculate and regulate power in a circuit to prevent damage to components.

## 5. What are some ways to increase or decrease power in a circuit?

There are a few ways to increase or decrease power in a circuit. One way is to change the voltage or current, as they are directly proportional to power. Another way is to add or remove components, which can affect the overall resistance of the circuit and therefore alter the power. Additionally, using different types of components with different power ratings can also impact the power in a circuit.

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