## Unit 9

1. In Tango and Cash Sylvester Stallone slid down a high voltage 1000 Volt wire. He put both of his hands on a high voltage wire. He didn't get electrocuted. Is this an example of "Phalse" physics? Explain.

No, this is perfectly sound physics. Since Stallone put both of his hands on the wire there was no potential difference and no charge flowed through him which allowed him to slide down the high voltage wire safely.

2. A charge of 1 Coulombs went through a light bulb in 2 seconds when it was placed across a potential difference of 120 Volts.

a. Determine the current that flowed through the light bulb.

b. What power rating does the light bulb have?

For challenges 3 & 4 use the diagram below.

__e_3. The amount of work done by a battery to move 3 Coulombs of charge between terminal A and terminal B is ______ Joules.

a. 4 b. 8 c. 12 d. 18 e. 36

__b _4. If conventional current flows from terminal B to terminal A, then:

a. terminal B is negative and terminal A is positive.

b. terminal B is positive and terminal A is negative.

c. terminal B and terminal A have the same potential.

_a. 5. The current in a circuit is:

a. like the water flow through a circuit.

b. like the water pressure across a circuit.

c. is the same as the voltage across a circuit.

__b._6. The voltage in a circuit is:

a. like the water flow through a circuit.

b. like the water pressure across a circuit.

c. is the same as the current running through a circuit.

__c__7. In your circuit lab you have been using round bulbs and long bulbs. You charged up a blue capacitor through each bulb. Charging the capacitor with the round bulbs, you observed the bulbs light brighter and for a shorter period of time as compared to charging the capacitor with the long bulbs. The best explanation of this is:

a. The capacitor stored more charge when using the round bulbs.

b. The capacitor stored less charge when using the round bulbs.

c. The capacitor stored the same charge when using the round bulbs as with the long bulbs, but allowed the charge to flow at a quicker rate because the round bulbs had less resistance.

d. The capacitor stored the same charge when using the round bulbs as with the long bulbs, but allowed the charge to flow at a quicker rate because the round bulbs had more resistance.

8. The circuit set up in your lab is shown below. You and your partner observe the compass in the circuit shown to the left deflects in a clockwise direction. You and your partner rotate the circuit as shown below.

In what direction will you see the compass rotate for the rotated circuit rotate. Explain.

Clockwise. The same current will be flowing through the compass. It does not matter where between the two resistors the compass is placed since the wire has negligible resistance. ?????????

9. Draw the direction of conventional current flow on each of the situations shown below. On the first circuit, the capacitor is charging. On the second circuit the capacitor is uncharging.

Left: From the positive charge to the negative charge.

Right: From the negtive charge to the positive.

10. A current is traveling through a wire with 100 ½ of resistance. If the current traveling through the wire was 0.3 Amps, determine the voltage (potential difference or potential drop) across the wire.

V=(.3 amps)(100 ½)=30 V

11. The current through a light bulb connected across a 120 Volt outlet is 0.5 A. At what rate does the bulb convert electrical energy to light and thermal energy?

P=(.5 amps)(120 V)=60 W

12. A car battery causes a current of 2.0 A to flow through a lamp while 12 V is across it. Determine the power used by the lamp.

P=(2 amps)(12 V)=24 W

13. A 75 watt lightbulb is connected to a 120 V outlet

a. Determine the current through the wire.

Analyze the following circuits following the five step plan outlined in example 6 above.

15. 5 ½, 10 ½, and 15 ½ resistors are connected in series across a 12 Volt battery.

a. Determine the total resistance of the circuit.

R=5 ½+10 ½+15 ½=30 ½

b. Determine the current through the 10 ½ resistor.

c. Determine the potential drop across the 5 ½, 10½ and 15 ½ resistors.

d. Which resistor had the largest potential drop across it? Explain.

The 15 ½ resistor caused the largest potential drop since it has the highest resistivity ot the electricity.

16. 50 ½, 90 ½, and 150 ½ resistors are connected in series across a 50 Volt source.

a. Determine the total resistance of the circuit.

R=50 ½+90 ½+150 ½=290 ½

b. Determine the current through the 150 ½ resistor.

Current is always the same in one wire so the enitre resistivity of thw wire must be used to find the current through any of the resistors.

c. Determine the potential drop across the 90 ½ resistor.

d. Determine the power dissipated by each resistor.

e. Determine the total power dissipated by the 50 V source.

f. How does the total power dissipated by the 50 V source compare the sum of the powers dissipated by each resistor? Explain the relationship.

They are the same.

Analyze the following circuits following the five step plan outlined in example 7 above.

17. 5 ½, 10 ½, and 15 ½ resistors are connected in parallel across a 12 Volt battery.

a. Determine the total resistance of the circuit.

b. Determine the potential drop across the 5 ½, 10½ and 15 ½ resistors.

12 V for all of them since they all should be equivalent to the total potential drop of the system.

c. Determine the current through the 10 ½ resistor.

d. Which resistor had the largest current through it? Explain.

The 5 ½ resistor had the largest current through it because the only non-constant in the equation is resistance and since it it on the bottom of the fraction, the current is inversely proportional to it .

18. 50 ½, 90 ½, and 150 ½ resistors are connected in parallel across a 50 Volt source.

a. Determine the total resistance of the circuit.

b. Determine the potential drop across the 150 ½ resistor.

50 V

c. Determine the current through the 90 ½ resistor.

50 V

d. Determine the power dissipated by each resistor.

e. Determine the total power dissipated by the 50 V source.

f. How does the total power dissipated by the 50 V source compare the sum of the powers dissipated by each resistor? Explain the relationship.

They are equivalent.

19. A 40 ½ resistor is connected in series with 70 ½, 120 ½, and 100 ½ resistors connected in parallel across a potential difference of 120 V. Determine:

a. The effective resistance of the circuit.

120 V

b. The potential drop across the 40 ½ resistor.

First, the resistance of the overall circuit must be calculated. To do this we must calculate the total resistance of the parallel part and add that with the 40 ½ of the straight series part. Then calculating current is easy since the current through the 40 ½ part is equal to the total current of the circuit. Once the current is discovered, the potential drop can be solved for using V=IR.

V=(1.70A)(40½)=68v

c. The potential drop across the 120 ½ resistor.

We know that the potential through each of the parallel wires will be equivalent to the potential drop through all of them and we know that the potential drop through 40 ½ resistor and the potential drop through the parallel wires should add up to the potential drop through the entire circuit.

120v=68v+v

v=62

d. The current through the 40 ½ resistor.

I=1.70A (See part b.)

e. The current through the 70 ½ and 100 ½ resistors.

v=62 for the all of the parallel wires.

20. 80 ½ and 40 ½ resistors in connected in series. These two resistors are in turn connected in parallel with 30 ½ and 50 ½ resistors which are connected across a 120 V potential difference. Determine:

a. The effective resistance of the circuit.

b. The potential on the 30 ½ resistor.

120v

c. The potential drop on the 80 ½ and 40 ½ resistors.

We must determine the current through this section of parallel wires.

Then we can just plug into V=IR

V=(1A)(40½)=40v

V=(1A)(80½)=80v

d. The current through the 80 ½ and 40 ½ resistors.

1A. (see c)

e. The current through the 50 ½ resistor.

21.

R1 10½ 29.2v 2.92

R2 20½ 58.4v 2.92

R3 30½ 32.4v 1.08

R4 40½ 32.4v 0.81

R5 50½ 14.7v 0.29

R6 60½ 17.7v 0.29

R7 70½ 32.4v 0.46

R8 80½ 20.4v 0.25

R9 90½ 12.0v 0.13

R10 100½ 12.0v 0.12

R total 41.2½ 120v 2.92

Glenbrook South Physics Team - Tom Henderson, John Lewis, Neil Schmidgall, Dave Smith, Suzanne Webb & Brian Wegley
4000 West Lake Ave
Glenview, IL 60025 - 1200
(847)729-2000
Page Maintained by: Brian Wegley

Last Updated: October 8, 1997