# Simple power equation question

Suppose I have a circuit with voltage 4V and current 1A. By $P=I~V$ the total power across the circuit is 4W.

I have a $20~\Omega$ resistor and want to calculate the power loss over the resistor. By $I^2 R$ that is 20W.

This doesn't make sense to me. If the total power across the circuit is 4W, how can I lose 20W of power over the resistor. Where is my understanding wrong here?

• The current flowing through the resistor would be 0.2 A(V / R = I ), so the power loss across it would be 0.8 W. "Suppose I have a circuit with voltage 4V and current 1A" are you describing a power supply with this statement. May be you should provide a schematic. – karthik Jay Aug 30 '17 at 4:07
• A $20\:\Omega$ resistor would require $20\:\textrm{V}$ to cause $1\:\textrm{A}$ through it. So your voltage is wrong. You don't get to pick parts from circuit 1 and different parts from circuit 2 and ask "why?" – jonk Aug 30 '17 at 4:53
• @karthikJay yes I'm describing the output of my power supply – user Aug 30 '17 at 5:09
• What do you mean "you are describing the output of your power supply"? Are you describing what is written on a PS label, or do you have a more sophisticated PS with DMM displays, or are you measuring the output with a Voltmeter and a separate Ampermeter? – Ale..chenski Aug 30 '17 at 5:52
• If your power supply can deliver UP TO 1 Amp at 4 volts, this does not mean that it WILL deliver 1 Amp. The current drawn from the power supply will be determined by the load resistance, provided that the supply can produce the required current. A power supply does not force its advertised current through the load. – Peter Bennett Aug 30 '17 at 6:18

## 1 Answer

If the total power across the circuit is 4W, how can I lose 20W of power over the resistor

You cannot get 1A through a 20Ω resistor with only 4W, so your assumption that it will draw 1A is wrong. If the voltage across it is 4V then it will draw 4V/20Ω = 0.2A, and dissipate 4V * 0.2A = 0.8W.

If the total power absorbed by the circuit is 4W then either there are other components drawing current in parallel to make the total 1A, or the voltage is being boosted before being put across the resistor. If the resistor is the only component then the current must actually be 0.2A, not 1A.