Last Regular Homework of the Fall Semester (Next week will be a full semester review)

Ohm's Law

(Simple)

1. What potential difference is required to produce a 35 mA current in a 47 Ω resistance?

∆V = I•R = .035 A • 47 Ω = 1.645 V

2. What current is produced by a 12 V potential difference in a 68 Ω resistance?

I = ∆V÷R = 12 V ÷ 68 Ω = .176 A

3. What resistance is required where a 6 V battery is to deliver a 45 mA current?

R = ∆V/I = 6 V ÷ .045 A = 133 Ω

(More Interesting)

For the following questions, three resistors, of 5, 12 and 15 ohms are networked and connected to a 12 V battery.

4. If the resistors are in series:

a. what is the total resistance?

5 Ω + 12 Ω + 15 Ω = 32 Ω

b. What is the current through each?

The current through each in a series is the same as the total current: I = ∆V ÷ R = 12 V ÷ 32 Ω = .375 A

c. What is the potential difference across each?

5 Ω : I•R = .375 A • 5 Ω = 1.875 V, 12 Ω = 4.5 V, 15 Ω = 5.625 V

d. How much power is delivered to each?

P = I∆V = .375 • 1.875 V = .703 W, 1.69 W, 2.11 W

e. Explain how the loop rule applies.

The sum of the ∆V s across the resistors add to the battery to give zero: -1.875 + - 4.5 + - 5.625 + 12 = 0

5. If the resistors are in parallel:

a. what is the total resistance?

1/5 + 1/12 + 1/15 = 1/R

R = 2.86 Ω

b. What is the current through each?

I = ∆V/R = 12/5 = 2.4 A and 12/12 = 1 A and 12 /15 = .8 A

c. What is the potential difference across each?

12 V

d. How much power is delivered to each?

P = I∆V = 2.4 • 12 = 28.8 W, 12 W and 9.6 W

e. Explain how the junction rule applies.

The total current: 12/2.86 = 4.2 A = 2.4 A+ 1 A+ 0.8 A

6. The 5 Ω resistor is in series with the combination of the other two, in parallel with each other:

a. What is the total resistance (hint: solve the parallel part first)

1/R = 1/12 + 1/15 so R = 6.7 Ω + 5 Ω = 11.7 Ω

b. What is the potential difference across the 5 Ω resistance?

I = 12 V/11.7 Ω = 1.03 A; ∆V = I•R = 1.03 A • 5 Ω = 5.15 V

c. What is the current through the 15 ohm resistance?

∆V = 12 - 5.15 = 6.85 V; I = ∆V/R = 6.85 V ÷ 15 Ω = .457 A

d. Explain how the loop rule applies.

+12 + -5.15 + - 6.85 = 0 or -6.85 + - 6.85 = 0 (I'll explain in class)

e. Explain how the junction rule applies.

I through the 5 = 12V ÷ 11.7 Ω = 1.03 A; 1.03 A= .457 A+ .573 A

7. The 5 Ω and 12 Ω resistances are in series. That combination is then placed in parallel with the 15 Ω resistance.

a. What is the potential difference across the 15 Ω resistance?

12 V

b. What is the total resistance of the network?

5 + 12 = 17 Ω then, 1/15 + 1/17 = 1/R so R = 7.97 Ω

c. What is the total current in the network?

I = ∆V/R = 12 ÷ 7.97 Ω = 1.51 A

d. Explain how the loop rule applies.

The sum of the changes in potential across the 5 and 12 Ω resistors must add to - 12. The change in potential across the 15 Ω resistor must itself be - 12 V.