## SAT Physics Subject Test

**Chapter 10 ****Direct Current Circuits**

**ELECTRIC CIRCUITS**

An electric current is maintained when the terminals of a voltage source (a battery, for example) are connected by a conducting pathway, in what’s called a **circuit**. If the current always travels in the same direction through the pathway, it’s called a **direct current**.

The job of the voltage source is to provide a potential difference called an **electromotive** **force**, or **emf**, which drives the flow of charge. The emf isn’t really a force; it’s the work done per unit charge, and it’s measured in volts.

To try to imagine what’s happening in a circuit in which a steady-state current is maintained, let’s follow one of the charge carriers that’s drifting through the pathway. (Remember, we’re pretending that the charge carriers are positive, just like we imagine that test charges in an electric field are positive.) The charge is introduced by the positive terminal of the battery and enters the wire, where it’s pushed by the electric field. It encounters resistance, bumping into the relatively stationary atoms that make up the metal’s lattice and setting them into greater motion. So the electrical potential energy that the charge had when it left the battery is turning into heat. By the time the charge reaches the negative terminal, all of its original electrical potential energy is lost. To keep the current going, the voltage source must do positive work on the charge, forcing it to move from the negative terminal toward the positive terminal. The charge is now ready to make another journey around the circuit.

**Energy and Power**

When a carrier of positive charge *q* drops by an amount *V* in potential, it loses potential energy in the amount *qV*. If this happens in time *t*, then the rate at which this energy is transformed is equal to (*qV*)/*t* = (*q*/*t*)*V*. But *q*/*t* is equal to the current, *I*, so the rate at which electrical energy is transferred is given by the equation

*P* = *IV*

This equation works for the power delivered by a battery to the circuit as well as for resistors. The power dissipated in a resistor, as electrical potential energy is turned into heat, is given by *P* = *IV*, but because of the relationship *V* = *IR*, we can express this in two other ways.

*P* = *IV* = *I*(*IR*) = *I*^{2}*R*

or

Resistors become hot when current passes through them; the thermal energy generated is called **joule heat**.