Question

A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V rms to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary

Answer 1

a. 41

b. i. 15 mA ii. 625 mA

c. 192 Ω

Step-by-step explanation:

Here is the complete question

A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V (rms) to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary? (b) What are the rms currents in the primary and secondary coils of the transformer? (c) What is the effective resistance that the 120-V source is subjected to?

Solution

(a) What is the ratio of the number of turns in the secondary to the number of turns in the primary?

For a transformer N₂/N₁ = V₂/V₁

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V and V₂ = voltage of secondary coil = 5000 V

So, N₂/N₁ = V₂/V₁

N₂/N₁ = 5000 V/120 V = 41.6 ≅ 41 (rounded down because we cannot have a decimal number of turns)

(b) What are the rms currents in the primary and secondary coils of the transformer?

i. The rms current in the secondary

We need to find the current in the secondary from

P = IV where P = power dissipated in secondary coil = 75.0 W, I =rms current in secondary coil and V = rms voltage in secondary coil = 5000 V

P = IV

I = P/V = 75.0 W/5000 V = 15 × 10⁻³ A = 15 mA

ii. The rms current in the primary

Since N₂/N₁ = V₂/V₁ = I₁/I₂

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V, V₂ = voltage of secondary coil = 5000 V, I₁ = current in primary coil and I₂ = current in secondary coil = 15 mA

So, V₂/V₁ = I₁/I₂

V₂I₂/V₁ = I₁

I₁ = V₂I₂/V₁

= P/V₁

= 75.0 W/120 V

= 0.625 A

= 625 mA

(c) What is the effective resistance that the 120-V source is subjected to?

Using V = IR where V = voltage = 120 V, I = current in primary = 0.625 A and R = resistance of primary coil

R = V/I

= 120 V/0.625 A

= 192 V/A

= 192 Ω