A series R-C circuit is connected to AC voltage source. Consider two cases; (A) when C is without a

Q: Why does the current increase when the dielectric is inserted?

Inserting a dielectric increases the capacitance, which decreases the capacitive reactance. A lower reactance means lower total impedance, allowing more current to flow from the AC source.

Q: Doesn't $V_C = I_R X_C$ mean $V_C$ could increase if $I_R$ increases?

It's a tug-of-war! $I_R$ increases, but $X_C$ decreases. To know which effect wins, we must substitute $I_R = V/Z$ and simplify the expression. The math shows that the decrease in $X_C$ dominates, so $V_C$ decreases.

Q: What exactly does the dielectric constant do?

The dielectric constant (or relative permittivity) is a measure of a material's ability to store electrical potential energy under the influence of an electric field. It multiplies the original capacitance by its value.

Q: Why do we use rms values for current and voltage here?

In AC circuits, unless specified otherwise, we always compare the rms (root mean square) values because they represent the effective DC-equivalent values for power and measurements.

Q: Would the voltage across the resistor change?

Yes! Since the current $I_R$ increases and the resistance $R$ is constant, the voltage across the resistor $V_R = I_R R$ will increase.

Question

A series $R$-$C$ circuit is connected to AC voltage source. Consider two cases; ($A$) when $C$ is without a dielectric medium and ($B$) when $C$ is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true? Options: (A) $I_R^A > I_R^B$ (B) $I_R^A < I_R^B$ (C) $V_C^A > V_C^B$ (D) $V_C^A < V_C^B$

Sigma Percentile · Accepted Answer

### The Setup: A Series R-C Circuit

Imagine you are standing in front of a simple yet elegant electrical circuit. We have a resistor $R$ and a capacitor $C$ connected in series to an alternating current (AC) voltage source.

I know this might look like a standard textbook problem, but let's take a breath and dive into the physics of what happens when we tweak just one component.

### The Concept of Impedance

In an AC circuit, the total opposition to the flow of current is called **impedance**, denoted by $Z$. For a series R-C circuit, the impedance is a combination of the resistance $R$ and the capacitive reactance $X_C$.

$$Z = \sqrt{R^2 + X_C^2}$$

The capacitive reactance $X_C$ depends inversely on the frequency $\omega$ and the capacitance $C$:

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

### Enter the Dielectric

Now, let's introduce a twist. In case (B), we fill the capacitor with a dielectric medium of constant $K = 4$.

What does this do? A dielectric material enhances the capacitor's ability to store charge, effectively multiplying its capacitance by the dielectric constant.

$$C' = 4C$$

So, the capacitance $C$ has increased significantly.

### The Domino Effect on Current

Let's trace the domino effect of this increased capacitance. Since $C$ is in the denominator of the reactance formula, a larger $C$ means a smaller $X_C$.

With a smaller $X_C$, the total impedance $Z$ of the circuit decreases.

According to Ohm's law for AC circuits, the current $I_R$ is inversely proportional to the impedance:

$$I_R = \frac{V}{Z}$$

Since the impedance $Z$ has dropped, the current $I_R$ must surge upwards. Therefore, the current in case (B) is greater than in case (A).

**Result 1:** $I_R^A < I_R^B$

### The Tug-of-War: Voltage Across the Capacitor

Now comes the tricky part. What happens to the voltage across the capacitor, $V_C$?

We know that $V_C = I_R X_C$.

Here is the catch: the current $I_R$ has increased, but the reactance $X_C$ has decreased. It's a mathematical tug-of-war! To find out which effect dominates, we need to combine our equations into a single, clear expression.

### The Mathematical Resolution

Let's substitute $I_R = \frac{V}{Z}$ into our voltage equation:

$$V_C = \left(\frac{V}{\sqrt{R^2 + X_C^2}}\right) X_C$$

To simplify this, let's bring $X_C$ into the square root in the denominator:

$$V_C = \frac{V}{\sqrt{\left(\frac{R}{X_C}\right)^2 + 1}}$$

Now, substitute $X_C = \frac{1}{\omega C}$ back into the equation:

$$V_C = \frac{V}{\sqrt{(\omega R C)^2 + 1}}$$

### Final Verdict

Look closely at this beautiful final expression. The capacitance $C$ now appears only in the denominator.

Since $C$ is larger in case (B), the entire denominator $\sqrt{(\omega R C)^2 + 1}$ becomes larger.

A larger denominator means the overall fraction becomes smaller. Thus, the voltage across the capacitor $V_C$ decreases when the dielectric is inserted.

**Result 2:** $V_C^A > V_C^B$

By carefully isolating the variable we changed, we resolved the tug-of-war and found the exact physical behavior of the circuit!

Select Answer:

Visualized Solution

$R$ - $C$ Circuit Setup

Impedance of $R$ - $C$ Circuit

Inserting the Dielectric

Effect on Impedance

Effect on Current

Voltage Across Capacitor

Analyzing the Denominator

Final Conclusion

The Sigma Insight: Alternating Current (AC) and Voltage

The Setup

The Concept of Impedance

Enter the Dielectric

The Domino Effect on Current

The Tug-of-War

The Mathematical Resolution

Final Verdict

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