X C

1. XC XL f fo Z R f fo

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

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

4. A copper coil is set just behind the wire loop of this circuit. As the switch is opened, • a clockwise current is induced in the copper loop. • a counter clockwise current is induced in the copper loop. • no current appears in the copper loop.

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

6. 1 Amp 120 V 120 V 240 V What is the voltage across the light bulb? 1) 60 V 2) 120 V 3) 240 V

7. 1 Amp 120 V 120 V 240 V 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

8. 1 Amp 120 V 120 V 240 V What is the voltage across the light bulb? A) 60 V B) 120 V C) 240 V Twice as many loops voltage doubles. THEN: half as many loops, voltage halves What is the current through the light bulb? A) 1/2 Amp B) 1 Amp C) 2 Amp s Power in = Power out Voltage out = 120 V 240 V  1 Amp = 120 V  I so I = 2 Amp

9. 2) a counter clockwise current is induced in the copper loop. • A clockwise current is created in the black wire loop, establishing a sudden magnetic • field pointing down into the center of the copper coil. To counter this, the coil will • try to create a field of its own pointing out. It does this with a counter clockwise current. • 2) a counter clockwise current is induced in the copper loop. • While current flows the wire loop’s field points into the screen within its center, but • doubles back outside the loop. That means where the copper coil lies, the field initially • points out. When the switch is opened, this field suddenly vanishes. To counter the • disappearance of flux, the coil tried to create its own, which requires a clockwise • induced current. • a clockwise current is induced in the copper loop. • The coil tried to replace the flux that starts to disappear as soon as the switch is opened. • less than 6 V The voltage across B is zero only a changing magnetic flux induces • an EMF. Batteries provide DC current. • 2) 120 V Twice as many loops voltage doubles. THEN: half as many loops, voltage halves • 3)2 Amps Power in = Power out, Voltage out = 120 V • 240 V  1 Amp = 120 V  I so I = 2 Amp