Mastering Physics Solutions: Exercise 20.30

Exercise 20.30

Part A = 20 V Click to use the calculator/solver for this part of the problem
Part B = 4200 turns Click to use the calculator/solver for this part of the problem
Part C = 1.6 A Click to use the calculator/solver for this part of the problem

Solutions Below:

The primary coil of an ideal transformer is connected to a 120-V source and draws 1.0 A. The secondary coil has 700 turns and supplies an output current of 6.0 A to run an electrical device.

Part A

What is the voltage across the secondary coil?
Express your answer using two significant figures.

The current transformation ratio for a transformer is equal to Nprimary / Nsecondary, where N is the number of turns (in the primary and secondary windings, respectively). The voltage transformation ratio is the opposite (Nsecondary / Nprimary). Note also that the voltage and ratios are the ratios of secondary to primary voltage or current. Since this problem gives us the current in both windings, start from there:

CurrentRatio = Nprimary / Nsecondary
(6 / 1) = Nprimary / 700
Nprimary = 4200

Since we have both the primary and secondary turns now, we can solve for the voltage ratio and determine the voltage across the secondary voltage:

VoltageRatio = Nsecondary / Nprimary
Vsecondary / 120 = 700 / 4200
Vsecondary = 20 V

20 V

Part B

How many turns are in the primary coil?
Express your answer as an integer.

Just use the same formula as in Part A:

CurrentRatio = Nprimary / Nsecondary
(6 / 1) = Nprimary / 700
Nprimary = 4200

4200 turns

Part C

If the maximum power allowed by the device (before it is destroyed) is 290 W, what is the maximum input current to this transformer?
Express your answer using two significant figures.

We need to find the load resistance in the transformer to solve this problem. The load resistance is just the resistance in the secondary winding. We know both the secondary voltage and current from Part A:

R = Vsecondary / Isecondary
R = 20 / 6
R = 3.333

Now we can plug this into the formula for power. Start by finding the maximum secondary current:

W = (Isecondary)2 * R
290 = (Isecondary)2 * 3.33
87 = (Isecondary)2
Isecondary = 9.33

Since the current transformation ratio is 6, just divide by 6:

9.33 / 6 = 1.56

It’s important to find the load resistance in order to correctly solve this problem. In case you try to solve by the formula W = V * I, you’ll find that you get an incorrect answer.
1.6 A

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