Statement I: To get a steady DC output from the pulsating voltage received from a full wave rectifie

Q: Why does a capacitor block DC?

The opposition offered by a capacitor is called capacitive reactance, $X_C = \frac{1}{2\pi f C}$. Since the frequency $f$ of Direct Current (DC) is zero, the reactance becomes infinite. Therefore, it acts as an open circuit for DC.

Q: Why must the capacitor be connected in parallel and not in series?

If connected in series, the capacitor would block the DC from reaching the load entirely! By connecting it in parallel, it acts as a 'drain' or bypass for the AC ripples, allowing the blocked DC to continue straight into the load.

Q: Why is the inductor connected in series?

An inductor allows DC to pass ($X_L = 0$) but blocks AC ($X_L = 2\pi f L$). By placing it in series with the load, it acts as a gatekeeper, stopping the AC ripples from passing through while letting the DC flow freely to the load.

Q: Can we use both a capacitor and an inductor together?

Absolutely! In fact, practical power supplies use combinations of both, such as LC filters or $\pi$-filters. The inductor chokes the AC in series, and the capacitor bypasses any remaining AC in parallel, resulting in an incredibly smooth DC output.

Q: What exactly is a 'ripple'?

A ripple is the residual periodic variation (the AC component) of the DC voltage output of a power supply derived from an AC source. It's the 'waviness' you see on top of the DC signal before it is properly filtered.

Question

Statement I: To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load $R_L$.
Statement II: To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with $R_L$.
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Statement I is true but statement II is false.

(B) Statement I is false but statement II is true.

(C) Both statement I and statement II are false.

(D) Both statement I and statement II are true.

Sigma Percentile · Accepted Answer

### The Quest for Pure DC

When we use a full-wave rectifier to convert Alternating Current (AC) into Direct Current (DC), the job is only half done. The output we receive is not the perfectly flat, steady DC line we desire; instead, it is a **pulsating DC**. This pulsating output contains the desired DC component along with unwanted AC fluctuations, commonly referred to as **ripples**.

To eliminate these ripples and extract a smooth, steady DC voltage for our electronic devices, we employ circuits known as **filters**. Let's explore the two most fundamental types of filters mentioned in the problem.

### The Capacitor Filter (Parallel Connection)

Statement I proposes connecting a capacitor in parallel across the load resistance $R_L$. To understand why this works, we must look at the capacitive reactance, given by the formula:

$$X_C = \frac{1}{\omega C}$$

For the DC component, the frequency is zero ($f = 0$), which means the angular frequency $\omega = 0$. Consequently, the reactance $X_C$ becomes infinite. The capacitor acts as an open circuit and completely **blocks DC**.

However, for the AC ripples, $\omega$ is non-zero, making $X_C$ a finite, relatively low value. The capacitor provides a low-resistance bypass path for these ripples. As a result, the AC components flow through the capacitor, bypassing the load, while the pure DC component is forced to flow entirely through the load resistance $R_L$. Thus, Statement I is perfectly true.

### The Inductor Filter (Series Connection)

Statement II suggests connecting an inductor in series with the load resistance $R_L$. The behavior of an inductor is governed by its inductive reactance:

$$X_L = \omega L$$

An inductor inherently opposes any change in current. For the steady DC component, $\omega = 0$, which means $X_L = 0$. The inductor acts as a simple wire, allowing the DC to pass through without any opposition.

Conversely, for the high-frequency AC ripples, $\omega$ is large, resulting in a high inductive reactance $X_L$. The inductor offers massive resistance to the AC components, effectively **choking** them and preventing them from reaching the load. Therefore, only the steady DC makes it to the output. This confirms that Statement II is also entirely true.

### Conclusion

Both the parallel capacitor and the series inductor are highly effective, fundamental techniques for filtering pulsating DC into steady DC. In practical applications, they are often combined into more complex configurations (like L-section or $\pi$-filters) for even better smoothing. Since both statements accurately describe valid filtering methods, the correct choice is that both statements are true.

Select Answer:

Visualized Solution

Analyzing Statement I: Capacitor Filter

Analyzing Statement II: Inductor Filter

Final Conclusion

The Sigma Insight: P-N Junction Diode

The Quest for Pure DC

The Capacitor Filter (Parallel Connection)

The Inductor Filter (Series Connection)

Conclusion

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