Problem 24 - Applied Physics

Question

A step-up transformer has \(30 \overline{0} 0\) turns in the secondary coil and \(20 \overline{0}\) turns in the primary coil. The primary is supplied with alternating current with an effective voltage of \(90.0 \mathrm{~V}\). (a) Find the voltage in the secondary coil. (b) If the current in the secondary coil is \(2.00 \mathrm{~A}\), what current flows in the primary coil? (c) What power is developed in the primary coil? (d) What power is developed in the secondary coil?

Step-by-Step Solution

Answer

The secondary voltage is 1350 V, primary current is 30.0 A, primary power is 2700 W, and secondary power is 2700 W.

Step 1: Identify the Transformer Formula

The voltage in the secondary coil \( V_s \) can be calculated using the transformer formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), where \( V_p = 90.0 \text{ V} \) is the primary voltage, \( N_s = 3000 \) is the number of turns in the secondary, and \( N_p = 200 \) is the number of turns in the primary.

Step 2: Solve for Secondary Voltage

Rearrange the transformer formula to calculate the secondary voltage: \( V_s = V_p \cdot \frac{N_s}{N_p} = 90.0 \cdot \frac{3000}{200} = 1350 \text{ V} \).

Step 3: Calculate Current in Primary Coil

Use the power equivalence in a transformer: \( V_p \cdot I_p = V_s \cdot I_s \). We know \( V_s = 1350 \text{ V} \) and \( I_s = 2.00 \text{ A} \), substitute these into the equation to solve for \( I_p \): \( 90.0 \cdot I_p = 1350 \cdot 2.00 \), which simplifies to \( I_p = \frac{2700}{90.0} = 30.0 \text{ A} \).

Step 4: Determine Power in Primary Coil

Calculate the power in the primary coil using the formula \( P_p = V_p \cdot I_p = 90.0 \times 30.0 = 2700 \text{ W} \).

Step 5: Determine Power in Secondary Coil

Calculate the power in the secondary coil using \( P_s = V_s \cdot I_s = 1350 \times 2.00 = 2700 \text{ W} \).

Key Concepts

Transformer Formula

To understand how a transformer functions, it's crucial to grasp the transformer formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). This formula shows the relation between the voltages and the number of turns in the secondary and primary coils of a transformer.

• \( V_s \) is the voltage in the secondary coil.
• \( V_p \) is the voltage in the primary coil.
• \( N_s \) is the number of turns in the secondary coil.
• \( N_p \) is the number of turns in the primary coil.

The equation implies that the voltage ratio of the secondary coil to the primary coil is equal to the ratio of the number of turns in the secondary coil to the primary coil.
This is how a transformer can "step-up" voltage, as seen in the exercise where the secondary coil voltage is increased due to a higher number of turns.

Secondary Coil

The secondary coil in a transformer is essential for transforming voltage levels from the primary coil. It is designed differently to manage either an increase or a decrease in voltage. In a step-up transformer like this one, the secondary coil has more turns than the primary coil.
This always results in a higher secondary voltage compared to the primary voltage, thanks to its higher turn count. In our example, the secondary coil has 3000 turns, which is much greater than the primary coil's 200 turns.
This considerable difference allows for the voltage to be "stepped-up" from 90.0 V to 1350 V. Understanding this role of the secondary coil is pivotal to comprehending how transformers adjust voltage levels in electrical circuits.

Primary Coil

The primary coil of a transformer is where it all begins. It receives the initial input of electrical energy. In this context, the primary coil is supplied with 90.0 V of alternating current. The main function of the primary coil is to initiate the electromagnetic induction process.
With 200 turns, this coil is smaller compared to the secondary coil, deliberately designed for the step-up transformer to increase voltage.
Without the primary coil's specific design and turns, the desired voltage transformation would not occur, making it a critical component in the transformer's function.

Power Calculation

Power in a transformer is an essential consideration for efficient energy transfer. In electrical terms, power (\( P \) ), is calculated through the formula \( P = V \cdot I \), where \( V \) is the voltage and \( I \) is the current.
In our example, power developed in both primary and secondary coils can be calculated.

• For the primary coil: \( P_p = V_p \times I_p = 90.0 \times 30.0 = 2700 \) W.
• For the secondary coil: \( P_s = V_s \times I_s = 1350 \times 2.00 = 2700 \) W.

Despite voltage transformation, the power remains equal between primary and secondary sides of an ideal transformer. This is key for understanding energy conservation within transformers, showing that energy is only converted, not lost.