A common transistor radio set requires 12 V (DC) for its operation. The DC source is constructed by

Q: Why do we completely ignore the rectifier in our calculation?

The rectifier's job is to convert AC to DC *after* the transformer has already stepped down the voltage. The transformer equation $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ strictly applies to the AC voltages across the primary and secondary coils. The rectifier doesn't change the physics of the transformer itself.

Q: What if the radio needed AC instead of DC? Would the answer change?

No, the answer would be exactly the same. If the radio ran on $12\text{ V}$ AC, we simply wouldn't need the rectifier. The transformer would still need to step down $220\text{ V}$ AC to $12\text{ V}$ AC, requiring the exact same turns ratio.

Q: Can we use a DC supply on the primary coil of the transformer?

Absolutely not! Transformers rely on Faraday's Law of Induction, which requires a *changing* magnetic field to induce a voltage in the secondary coil. A steady DC supply creates a static magnetic field, meaning no voltage would be induced, and the transformer would likely overheat and burn out.

Q: Why is $N_p$ so much larger than $N_s$?

Because this is a step-down transformer. To reduce the voltage from $220\text{ V}$ to $12\text{ V}$, the secondary coil must have proportionally fewer turns than the primary coil. The voltage is 'stepped down' by the exact same factor as the turns.

Q: What would happen if we accidentally connected the $220\text{ V}$ supply to the $24$-turn coil?

It would act as a massive step-up transformer! The voltage on the other side would be stepped up by a factor of $\frac{440}{24}$, creating a dangerously high voltage. Furthermore, the $24$-turn coil likely has low resistance and would draw a massive current, instantly burning out.

Question

A common transistor radio set requires $12	ext{ V (DC)}$ for its operation. The DC source is constructed by using a transformer and a rectifier circuit, which are operated at $220	ext{ V (AC)}$ on standard domestic AC supply. The number of turns of secondary coil are $24$, then the number of turns of primary are ............ .

Sigma Percentile · Accepted Answer

### The Wall Socket vs. The Radio

Imagine you want to listen to your favorite radio station. Your radio is a delicate electronic device that requires a gentle $12	ext{ V}$ Direct Current (DC) to operate. However, the power coming out of your wall socket is a roaring $220	ext{ V}$ Alternating Current (AC). If you were to connect the radio directly to the wall, it would instantly fry!

To bridge this massive gap, we need two distinct components: a **transformer** to step down the high voltage, and a **rectifier** to convert the AC into DC. The beauty of this problem is that we can isolate the transformer's job completely from the rectifier's job. The transformer only cares about AC voltages and the number of turns in its coils.

### The Transformer's Secret: The Turns Ratio

A transformer works on the principle of electromagnetic induction. It consists of two coils—the primary (connected to the wall) and the secondary (connected to the radio's rectifier)—wrapped around a common iron core.

The fundamental law governing ideal transformers states that the ratio of the voltages across the coils is directly proportional to the ratio of the number of turns in those coils. Mathematically, this is expressed as:

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

Where:
- $V_p$ is the primary voltage ($220	ext{ V}$)
- $V_s$ is the secondary voltage ($12	ext{ V}$)
- $N_p$ is the number of turns in the primary coil (what we need to find)
- $N_s$ is the number of turns in the secondary coil ($24$)

### Crunching the Numbers

Now, let's substitute our known values into the master equation. We want to find out how many turns the primary coil needs to successfully step down the voltage from $220	ext{ V}$ to $12	ext{ V}$, given that the secondary coil has $24$ turns.

$$ \frac{220}{12} = \frac{N_p}{24} $$

To isolate $N_p$, we multiply both sides of the equation by $24$:

$$ N_p = \frac{220 	imes 24}{12} $$

Notice how elegantly the numbers simplify. The fraction $\frac{24}{12}$ reduces perfectly to $2$.

$$ N_p = 220 	imes 2 $$

$$ N_p = 440 $$

### What About the Rectifier?

You might be wondering, "Why didn't we use the fact that the radio needs DC?" The answer lies in the sequence of operations. The transformer first takes the $220	ext{ V}$ AC and steps it down to $12	ext{ V}$ AC. The transformer's job is now done.

After the transformer, the rectifier circuit takes over. It receives the $12	ext{ V}$ AC and converts it into the $12	ext{ V}$ DC that the radio actually uses. Because the transformer equation only deals with the AC voltages across its own coils, the presence of the rectifier downstream does not alter our calculation for the number of turns.

Our final answer makes perfect physical sense: to step down the voltage significantly, the primary coil must have many more turns ($440$) than the secondary coil ($24$).

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{The Transformer Equation}

\text{Substituting Values}

\text{Solving for } N_p

\text{Final Calculation}

\text{Concept Check}

The Sigma Insight: Alternating Current (AC) and Voltage

The Wall Socket vs

The Transformer's Secret

Crunching the Numbers

What About the Rectifier?

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