Section 8.1Electric Charge and Current

1. (Exercises 3 and 4) How much electric current flows in a resistor through which 25 C of charge passes in 8.0 s?

According to the definition of electric current,

I = q / t = 25 C / 8.0 s = 3.1 A

Section 8.2Voltage and Electrical Power

2. (Exercises 9 and 10) What is the resistance of the circuit in paired exercise #1 in this study guide, if it is connected to a 13.6-V battery when a current of 3.1 A is flowing through it?

Using Ohm's law,

R = V / I = 13.6 V / 3.1 A = 4.4 ohms

3. (Exercises 21 and 22) Given three resistors with values of 5.0 ohms, 15 ohms, and 30 ohms, respectively:

a. What is their equivalent resistance if they are connected in parallel?

b. What is their equivalent resistance if they are connected in series?

a. In parallel, we can solve using the lowest common denominator method:

1/R_p = 1/R₁ + 1/R₂ + 1/R₃ = 1/5.0 ohms + 1/15 ohms + 1/30 ohms

1/R_p = 6/30 ohms + 2/30 ohms + 1/30 ohms = 9/30 ohms

Then we must take the reciprocal of both sides to get:

R_p = 30 ohms / 9 = 3.3 ohms

We could also have used the decimal method.

1/R_p = 1/R₁ + 1/R₂ + 1/R₃ = 1/5.0 ohms + 1/15 ohms + 1/30 ohms

R_p = 0.200/ohms + 0.067/ohms + 0.033/ohms = 0.30/ohms

Again we must take the reciprocal of both sides of the equation.

R_p = 1 ohms/ 0.30 = 3.3 ohms (the same result as above)

b. In series, the calculation is much easier because the values simply add together.

R_s = R₁ + R₂ + R₃ = 5.0 ohms + 15 ohms + 30 ohms = 50 ohms

Section 8.5Electromagnetism

4. (Exercises 23 and 24) A transformer supplying a voltage of 6.00 V when connected to a 110-V input line must handle a current of 3.50 A. What will be the maximum current that will flow in the primary windings of this transformer?

We know that if this is a good transformer with little power loss, the power in the primary must equal the power in the secondary.

P₁ = P₂ = V₁ I₁ = V₂ I₂

I₁ = V₂ I₂ / V₁ = 6.00 V ( 3.50 A ) / 110 V

I₁= 0.191 A

5. (Exercises 25 and 26) A color television set similar to the one in most students' homes requires a voltage of about 20,000 V ac to run the electron guns that produce a picture on the screen of the set. How many turns must be on the secondary winding of the transformer used to produce this voltage if there are 33 turns on its primary and the transformer is connected to a 110 V ac electric plug?

Here the transformer is of the step-up type and will have a turns ratio equal to:

V₂ / V₁ = N₂ / N₁

This means that:

N₂ = N₁ ( V₂ / V₁ ) = 33 turns ( 20,000 V / 110 V ) = 6000 turns

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