1. Exam 1Quiz 1Quiz 2

2. A series RLC circuit has a resistance of 25.0 , a capacitance of 50.0  F, and an inductance of 0.300 H. If driven by a 120v, 60Hz source, find: (a) the impedance of the circuit. (b) the current in the circuit. (c) the phase angle between current and voltage. (d) the rms voltage drop across each component. (e) the power dissipated in the circuit. (f) What f would make this circuit completely resistive? (g) What would the power dissipated in the resistor be then?

3. XCXC f XLXL f

4. XCXC f XLXL fofo f fofo R Z

5. A copper coil is set just behind the wire loop of this circuit. As the switch is closed, 1)a clockwise current is induced in the copper loop. 2)a counter clockwise current is induced in the copper loop. 3)no current appears in the copper loop.

6. A copper coil rests flat on a tabletop next to the wire loop of this simple circuit. As the switch is opened, 1)a clockwise current is induced in the copper loop. 2)a counter clockwise current is induced in the copper loop. 3)no current appears in the copper loop.

7. A copper coil is set just behind the wire loop of this circuit. As the switch is opened, 1)a clockwise current is induced in the copper loop. 2)a counter clockwise current is induced in the copper loop. 3)no current appears in the copper loop.

8. Compared to the voltage drop across coil A, the voltage across coil B is 1) less than 6 V 2) 6 V 3) greater than 6 V AB 6 V

9. Compared to the voltage drop across coil A, the voltage across coil B is 1) less than 6 V 2) 6 V 3) greater than 6 V AB 6 V The voltage across B is zero Only a changing magnetic flux induces an EMF. Batteries provide DC current.

10. 120 V 1 Amp What is the voltage across the light bulb? 1)60 V 2)120 V 3)240 V 240 V 120 V

11. 1 Amp What is the voltage across the light bulb? 1)60 V 2)120 V 3)240 V What is the current through the light bulb? 1)1/2 Amp 2)1 Amp 3)2 Amps 240 V 120 V

12. 1 Amp What is the voltage across the light bulb? A)60 V B)120 V C)240 V What is the current through the light bulb? A)1/2 Amp B)1 Amp C)2 Amps 240 V Twice as many loops voltage doubles. THEN: half as many loops, voltage halves Power in = Power out Voltage out = 120 V 240 V  1 Amp = 120 V  I so I = 2 Amp 120 V