Atomic Spectra Study with High-Voltage Discharge Tube: Let's Dive into the Exciting World of Physics!
How can a step-up transformer provide 5000 V rms to a high-voltage discharge tube while dissipating 75.0 W?
Options:
- a. 41
- b. i. 15 mA ii. 625 mA
- c. 192 Ω
Solution:
To provide 5000 V rms to the high-voltage discharge tube and dissipate 75.0 W, the step-up transformer uses a specific ratio of turns between the secondary and primary coils. Let's break down the process:
When a step-up transformer is connected to the line voltage (120 V rms) and designed to provide 5000 V rms to the discharge tube, the ratio of the number of turns in the secondary to the number of turns in the primary is calculated using the formula N₂/N₁ = V₂/V₁.
For this scenario, with V₁ = 120 V and V₂ = 5000 V: N₂/N₁ = 5000 V / 120 V ≈ 41 (rounded down)
Next, to determine the rms currents in the primary and secondary coils of the transformer, we utilize the power dissipated in the secondary coil (75.0 W) and the voltages involved:
i. The rms current in the secondary: Using P = IV and P = 75.0 W, V = 5000 V: I = P/V = 75.0 W / 5000 V = 15 mA
ii. The rms current in the primary: Calculating the current in the primary coil, we use the ratio equation N₂/N₁ = V₂/V₁ = I₁/I₂ and find: I₁ = V₂I₂ / V₁ = 75.0 W / 120 V = 0.625 A = 625 mA
Lastly, to determine the effective resistance that the 120-V source is subjected to, we apply Ohm's Law (V = IR) using the obtained current value:
R = V/I = 120 V / 0.625 A = 192 Ω