Question

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?
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?
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?
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.

Answer 1

Answer:

a) 0 V

b) 10 turns

c) 4000 turns

d) 12.5 A

e) 400 W

f) 0.5 A

g) 95.4%

Explanation:

A

0

B

To solve this, we would be using the simple relationship between voltage and number of turns

V1/V2 = N1/N2

500/25 = 200/N2

20 = 200/N2

N2 = 200/20

N2 = 10 turns

C

Here also, we would be using the relationship between current and the number of turns

I1/I2 = N2/N1

500/25 = N2/20

20 = N2/20

N2 = 20 * 20

N2 = 4000 turns

D

Like in the previous question, current and the number of turn relationship is used

N1/N2 = I2/I1

400/80 = I2/2.5

5 = I2/2.5

I2 = 5 * 2.5

I2 = 12.5 A

E

The power remains unchanged at 400 W

F

Power = Voltage * Current

P = VI

I = P/V

I = 60/120

I = 0.5 A

G

95.4%

Answer 2

The transformer is a device used to step up or step down voltage.

Part A;

Given that;

Es/Ep = Ns/Np

Es = voltage in the secondary coil

Ep = voltage in primary coil

Ns = Number of turns in secondary coil

Np = Number of coils in primary coil

Es = Ns/Np ×  Ep

Es = 200/100  × 1.5 V

Es = 3 V

Part B

Ns = Es/Ep × Np

Ns = 25/500 × 200

Ns = 10 turns

Part C

Ns/Np = Ip/Is

Ns = Ip/Is × Np

Ns = 500/25 × 200

Ns = 4000 turns

Part D

Ns/Np = Ip/Is

NsIs = NpIp

Is = NpIp/Ns

Is = 400 × 2.5/80

Is =12.5 A

Part E

The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.

Part F

Power supplied = 60 watts

Voltage of primary coil = 120 V

Since;

P = IV

I = P/V = 60/120 = 0.5 A

Part G

Since;

E = 100Pout/Pin

Pin = 120 V × 2 A = 240 W

Pout =  19.4 V ×  11.8 A = 228.92 W

E = 100(228.92/240)

E = 95.4%

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