Show your work and explain every step. Part A The primary coil of a transformer

Question

Show your work and explain every step.

Part A

The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?

The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?

Part B

A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.

Part C

A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.

Part D

In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current I2?

Express your answer numerically in amperes.

Part E

The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?

Part F

A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1 in the primary coil?

Express your answer in amperes.

Part G

The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency e of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: e=100Pout/Pin.

Express your answer as a percentage.

Explanation / Answer

A) np = 100 ns = 200 Vp = 1.5 v

np / ns = vp/vs

  hence vs = vp ( ns/np) = 1.5( 200/100) = 3 V

B) Vp = 500v vs = 25 v np = 200

  np / ns = vp/vs

ns = np ( Vs/vp) = 200 ( 25/500 ) = 10 turns

c) ip = 500A is = 25A   np = 200

ip/is = ns/np

ns = (ip/is) np = (500/25) 200 = 4000 turns

D) np = 400 ns = 80 ip = 2.5 A

   ip/is = ns/np

hence is = ip ( np/ns) = 2.5 ( 400/80) = 12.5 A

E) Power in the primary coil = power in the secondary coil

hence the power generated in secondary coil is also 400w

F) power P = 60W vp = 120V

ip = P/Vp = 60/120 = 0.5A

G) effieciency = (pout/ Pin) x100 =( VS is/ vpip) x100

= 19.4 x 11.8 x 100 / (120. x 2)

= 95.4%

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