Question

A transformer connected to a 120-V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA. (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 mA? (c) What current rating should the fuse in the primary circuit have?

Short answer

(a) 108.33 (b) 110.5 W (c) 0.92 A

Step-by-step solution

Identify Transformer Turn Ratio

The turn ratio of a transformer can be determined by the formula \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), where \( V_s \) and \( V_p \) are the secondary and primary voltages, and \( N_s \) and \( N_p \) are the secondary and primary turns, respectively. Plugging in the known values: \( V_s = 13,000 \, \text{V} \) and \( V_p = 120 \, \text{V} \), the equation becomes \( \frac{13,000}{120} = \frac{N_s}{N_p} \). Simplifying this, we find \( \frac{N_s}{N_p} = 108.33 \).

Calculate Power Supplied to Transformer

The power supplied to the transformer can be calculated using \( P = I_s \times V_s \), where \( I_s = 8.50 \, \text{mA} = 0.0085 \, \text{A} \) and \( V_s = 13,000 \, \text{V} \). Therefore, \( P = 0.0085 \, \text{A} \times 13,000 \, \text{V} = 110.5 \, \text{W} \).

Determine Primary Circuit Fuse Current Rating

The current in the primary circuit, \( I_p \), can be found using the conservation of power: \( P_p = P_s \), so \( V_p \times I_p = V_s \times I_s \). With \( V_p = 120 \, \text{V} \), \( V_s = 13,000 \, \text{V} \), and \( I_s = 0.0085 \, \text{A} \), we rearrange to find \( I_p = \frac{13,000 \, \text{V} \times 0.0085 \, \text{A}}{120 \, \text{V}} \). Simplifying gives \( I_p = 0.92 \, \text{A} \).

Key concepts

Turn Ratio

In any transformer, the turn ratio is essential because it links the voltages and the number of turns in the primary and secondary coils. The ratio of the number of turns is represented by the equation \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), where \( V_s \) is the secondary voltage, \( V_p \) is the primary voltage, \( N_s \) is the number of turns in the secondary, and \( N_p \) is the number of turns in the primary.
This equation tells us how the voltage is transformed from one coil to the next by adjusting the number of coil loops.
In this example, with a primary voltage of 120 V and a secondary voltage of 13,000 V, we found the turn ratio to be \( \frac{13,000}{120} = 108.33 \). This means the secondary has 108.33 times more turns than the primary, enabling it to increase the voltage from 120 V to 13,000 V effectively.
Understanding the turn ratio not only helps in designing transformers but also ensures they function properly, maintaining their expected voltage conversion.

RMS Voltage

Root Mean Square (RMS) voltage is crucial when dealing with alternating current (AC) circuits, like those in transformers. RMS voltage provides a measure of the equivalent DC voltage that delivers the same amount of power to a load. It is particularly important for ensuring devices receive the proper power levels without under- or overloading.
The RMS voltage is calculated by taking the square root of the average of the squares of all instantaneous voltages over one cycle.
In many household circuits, the RMS voltage is typically what is provided: in this exercise, the primary RMS voltage is 120 V and the secondary RMS voltage is 13,000 V.
Knowing and using RMS voltages allows electricians and engineers to work accurately and reliably with alternating current, guaranteeing that electrical appliances function properly within their specified limits.

Current Rating

Current rating is vital for determining the safety and efficiency of the electrical components within a transformer system. It specifies the maximum current that various parts of the circuit, including the wires and fuses, can safely carry.
This rating is crucial for preventing overheating and electrical fires.
In the context of the transformer in the original exercise, the secondary's current limit is set at 8.50 mA. Meanwhile, we calculate the primary circuit's current to ensure the connected components do not exceed their capability.
With a calculated primary current of 0.92 A, it is clear that this part of the circuit must have components rated for at least this current level for continued safe operation. Such precise calculations help prevent equipment damage and ensure operational efficiency.

Conservation of Power

The principle of conservation of power is central to understanding electrical transformers. It states that the power input into the transformer must equal the power output, modulo efficiency losses. This means that, ideally, what you put into the primary, you should get out of the secondary.
In mathematical terms, this is expressed as \( P_p = P_s \), or \( V_p \times I_p = V_s \times I_s \).
This principle keeps checks and balances on the power distribution across the transformer.
In the exercise, given a secondary output of 110.5 W, the primary must be similarly rated. This pushes the need to calculate the correct current rating for the primary circuit fuse (0.92 A).
Using this conservation law ensures transformers operate securely and correctly, highlighting the vital balance between input power and output power.