1. Figure 30.20A 5/2/161Oregon State University PH 213, Class #16

2. Figure 30.20B 5/2/162Oregon State University PH 213, Class #16

3. 5/2/16Oregon State University PH 213, Class #163 Ohm’s Law The “height” from which charge “falls” is a voltage difference,  V. The path by which charge falls is a conductor—any material that allows charge to flow freely. The rate at which an amount of charge, q, flows past any point in a conductor, q/  t, is called the electric current, I, and it’s measured in coulombs/sec, or Amperes (A). The ratio of the voltage difference,  V, to the resulting current, I, is called the resistance, R, a definition known as Ohm’s law:  V/I = Ror  V = IR R has units of ohms (  ).

4. A wire connects the positive and negative terminals of a battery. Two identical RESISTIVE wires connect the positive and negative terminals of an identical battery. Rank in order, from largest to smallest, the currents I a to I d at points a to d. A. I c = I d > I a > I b B. I a = I b > I c = I d C. I c = I d > I a = I b D. I a = I b = I c = I d E. I a > I b > I c = I d 5/2/164Oregon State University PH 213, Class #16

5. What is ∆V across the unspecified circuit element? Does the potential increase or decrease when traveling through this element in the direction assigned to I? A.∆V decreases by 2 V in the direction of I. B.∆V increases by 2 V in the direction of I. C.∆V decreases by 10 V in the direction of I. D.∆V increases by 10 V in the direction of I. E.∆V increases by 26 V in the direction of I. 5/2/165Oregon State University PH 213, Class #16

6. 5/2/16Oregon State University PH 213, Class #166 Resistors in Series A resistor is the reason for voltage drop in a circuit. It’s the site of “loss” of some of the potential energy of the charge flowing—tapped by some device for work or heat (or both). For every such resistor linked in series (one after another) in a circuit, the voltage drop across it is given by V = IR. Since the sum of all these voltage drops must be the total voltage drop, V T, around the circuit (all the way from the high side to the low side of the voltage source), this means: V = V T = V 1 + V 2 + V 3.... = IR 1 + IR 2 + IR 3... = I(R 1 + R 2 + R 3...) That is, we can replace all those individual resistors in series with a single resistor R S whose value is the sum of the individual R- values.

7. 5/2/16Oregon State University PH 213, Class #167 A 12-V car battery is connected to two lightbulbs in series. The resistance of the first bulb is 144 . The resistance of the second bulb is 72 . Using the low-voltage side of the battery as a 0-V reference, what is the voltage at the connection between the two resistors? 1. 4.00 V 2. 8.00 V 3. 12.0 V 4. 16.0 V 5. None of the above. (Always draw a picture of the circuit—it helps!)

8. 5/2/16Oregon State University PH 213, Class #168 Internal Resistance of a Voltage Source A battery or a generator or any other device that creates a constant voltage source has a little bit of resistance in its own materials—a value denoted as r. So the true voltage the device offers at its working terminals is usually slightly less than “rated.” How much less? That depends on the current it is delivering, because the voltage drop across any resistance is the product of the resistance value times the current flowing through that resistance. Example: A car battery has an internal resistance of about 0.01 . Note how that affects the terminal voltage if I = 10 A, or 100 A, etc.

9. 5/2/16Oregon State University PH 213, Class #169 The Energy Flow (Power) of an Electric Current The potential energy change of a given amount of charge, q, “falling” is given by:  U E = q  V Therefore, the rate at which that potential energy is converted (the power) is given by: P = q  V/  t And so, for a voltage difference (“height”) that doesn’t change with time, the steady power is given by: P =  V(q/  t) =  VI But since  V/R = I (that is,  V = IR), we have several ways to write the power of such a steady-state current flow: P =  V(I) P = (  V) 2 /R P = I 2 ·R

10. 5/2/16Oregon State University PH 213, Class #1610 You have two identical space heaters, each of which will produce 1000 W of heat output when plugged in to your wall outlet individually. But then you connect them in series and plug this combination into your wall outlet. What is their total power output now? 1. 500 W 2. 1000 W 3. 2000 W 4. 1414 W 5. None of the above. (Always draw a picture of the circuit—it helps!)