A mains transformer has a primary coil of 2500 turns and a secondary coil of 130 turns. The primary coil is connected to a mains supply where $V_{rms}$ is 230 V. The secondary coil is connected to a lamp of resistance 6.0 $ ext{Ω}$. The transformer is 100% efficient. What is the peak power dissipated in the lamp?

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A mains transformer has a primary coil of 2500 turns and a secondary coil of 130 turns - AQA - A-Level Physics - Question 25 - 2017 - Paper 2

Question 25

A mains transformer has a primary coil of 2500 turns and a secondary coil of 130 turns. The primary coil is connected to a mains supply where $V_{rms}$ is 230 V. The... show full transcript

Worked Solution & Example Answer:A mains transformer has a primary coil of 2500 turns and a secondary coil of 130 turns - AQA - A-Level Physics - Question 25 - 2017 - Paper 2

Step 1

Calculate the Secondary Voltage

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Answer

Using the transformer equation, we know that the voltage ratio is equal to the turns ratio:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

Where:

$V_p = 230 \text{ V}$ (primary voltage)
$N_p = 2500$ (number of turns in primary coil)
$N_s = 130$ (number of turns in secondary coil)

Rearranging gives us:

$V_s = V_p \times \frac{N_s}{N_p}$

Substituting in the values:

$V_s = 230 \times \frac{130}{2500} = 12.04 \text{ V}$

Step 2

Calculate the Power Dissipated in the Lamp

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Answer

The power dissipation in the lamp can be calculated using the formula:

$P = \frac{V_s^2}{R}$

Where:

$V_s = 12.04 \text{ V}$ (secondary voltage)
$R = 6.0 \text{ Ω}$ (resistance)

Substituting the secondary voltage:

$P = \frac{(12.04)^2}{6.0} \approx 24.20 \text{ W}$

Since we need to find the peak power dissipated, we can approximate this to the closest multiple of 12 W, yielding 24 W.

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