DC Circuits March 1, 2006

Power Source in a Circuit V The ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.

9. Consider the circuit shown in Figure P28.9. Find (a) the current in the 20.0-Ω resistor and (b) the potential difference between points a and b.

A REAL Power Source is NOT an ideal battery V E or E mf is an idealized device that does an amount of work E to move a unit charge from one side to another. By the way …. this is called a circuit! Internal Resistance

A Physical Battery Internal Resistance

Back to Potential Represents a charge in space Change in potential as one circuits this complete circuit is ZERO!

Consider a “circuit”. This trip around the circuit is the same as a path through space. THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!

To remember  In a real circuit, we can neglect the resistance of the wires compared to the resistors. We can therefore consider a wire in a circuit to be an equipotential – the change in potential over its length is slight compared to that in a resistor  A resistor allows current to flow from a high potential to a lower potential.  The energy needed to do this is supplied by the battery.

NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.  LOOP EQUATION The sum of the voltage drops (or rises) as one completely travels through a circuit loop is zero. Sometimes known as Kirchoff’s loop equation.  NODE EQUATION The sum of the currents entering (or leaving) a node in a circuit is ZERO

Take a trip around this circuit. Consider voltage DROPS : - E +ir +iR = 0 or E =ir + iR rise

Circuit Reduction i= E /R eq

Multiple Batteries

Reduction Computes i

Another Reduction Example PARALLEL

NOTICE ASSUMED DIRECTION OF TRAVEL Voltage Drops: - E 1 –i 1 R 1 + i 2 R 2 + E 2 -i 1 R 1 = 0 Second Loop -From “a” Let’s do it! NODE I 3 +i 2 = i 1

In the figure, all the resistors have a resistance of 4.0  and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R?

The Unthinkable ….

RC Circuit  Initially, no current through the circuit  Close switch at (a) and current begins to flow until the capacitor is fully charged.  If capacitor is charged and switch is switched to (b) discharge will follow.

Close the Switch I need to use E for E Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)

Really Close the Switch I need to use E for E Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)

This is a differential equation.  To solve we need what is called a particular solution as well as a general solution.  We often do this by creative “guessing” and then matching the guess to reality.  You may or may not have studied this topic … but you WILL!

Time Constant

Result q=CE(1-e -t/RC )

q=CE(1-e -t/RC ) and i=(CE/RC) e -t/RC

Discharging a Capacitor q initial =CE BIG SURPRISE! (Q=CV) i iR+q/C=0

In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy E th in the resistor as a function of time t. (a)What is the electric potential across the battery? (60) (b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39) (c) Did you put your name on your paper? (1) Looking at the graph, we see that the resistor dissipates 0.5 mJ in one second. Therefore, the POWER =i 2 R=0.5 mW

If the resistance is doubled what is the power dissipated by the circuit?