For a series L-C-R circuit with R = 100 ohms, L = 0.5 mH and C = 0.1 pF connected across 220 V-50 Hz

Q: Why do we ignore $X_L$ and $R$ when finding the phase angle?

Because $X_C$ is in the order of $10^{10} \Omega$, while $R$ is $100 \Omega$ and $X_L$ is less than $1 \Omega$. In the formula $\tan\phi = \frac{X_C - X_L}{R}$, subtracting $X_L$ from $X_C$ makes virtually no difference, and dividing by $R$ still leaves a massive number, making $\tan\phi \to \infty$ and $\phi \to 90^\circ$.

Q: What does it mean for a circuit to be "predominantly capacitive"?

It means the capacitor's opposition to alternating current is so much stronger than the inductor's or resistor's that the circuit behaves almost exactly like a pure capacitor connected to an AC source.

Q: Why is the capacitive reactance so high?

Capacitive reactance is inversely proportional to capacitance ($X_C = \frac{1}{\omega C}$). Since the capacitance is extremely small ($0.1 \text{ pF} = 10^{-13} \text{ F}$), the resulting reactance becomes astronomically large.

Q: Could this circuit ever be inductive?

Yes, if we drastically increased the frequency of the AC source. Since $X_L \propto f$ and $X_C \propto \frac{1}{f}$, at a high enough frequency, $X_L$ would surpass $X_C$.

Q: What is the significance of the phase angle being $90^\circ$?

A phase angle of $90^\circ$ indicates that the current leads the voltage by a quarter of a cycle. In a purely capacitive circuit, the power factor ($\cos 90^\circ$) is zero, meaning no real power is dissipated on average.

Question

For a series L-C-R circuit with $R = 100\Omega$, $L = 0.5	ext{ mH}$ and $C = 0.1	ext{ pF}$ connected across 220 V-50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is

Options:

(A) $0^\circ$, resistive circuit

(B) $\approx 90^\circ$, predominantly inductive circuit

(C) $0^\circ$ resonance circuit

(D) $\approx 90^\circ$, predominantly capacitive circuit

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a classic series L-C-R circuit. We are given a resistor $R = 100\,\Omega$, an inductor $L = 0.5\,	ext{mH}$, and a capacitor $C = 0.1\,	ext{pF}$. These components are connected in series across an alternating current source providing $220\,	ext{V}$ at a frequency of $50\,	ext{Hz}$.

Our goal is to determine the phase angle between the current and the supplied voltage, and to figure out the overall "nature" of the circuit. To do this, we need to calculate the opposition each component offers to the alternating current.

### The Inductive Reactance

Let's start with the inductor. The opposition it provides is called inductive reactance, denoted by $X_L$. The formula is:
$$X_L = \omega L = 2\pi f L$$

We substitute the given values, remembering to convert millihenries to henries:
$$X_L = 2\pi (50)(0.5 	imes 10^{-3})$$

Calculating this gives:
$$X_L = 50\pi 	imes 10^{-3}\,\Omega \approx 0.157\,\Omega$$

This is a remarkably small value. The inductor is barely putting up a fight against the $50\,	ext{Hz}$ current.

### The Capacitive Reactance

Now, let's turn our attention to the capacitor. Its opposition is the capacitive reactance, $X_C$, given by:
$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$

We substitute the frequency and the extremely small capacitance, converting picofarads to farads:
$$X_C = \frac{1}{2\pi (50)(0.1 	imes 10^{-12})} = \frac{1}{100\pi 	imes 10^{-13}}$$

Simplifying the denominator and bringing the power of ten to the numerator, we get:
$$X_C = \frac{10^{11}}{\pi}\,\Omega \approx 3.18 	imes 10^{10}\,\Omega$$

This is a massive, astronomical value! The capacitor is offering an immense opposition to the current.

### The Master Comparison

Let's compare the three players in our circuit:
- $R = 100\,\Omega$
- $X_L \approx 0.157\,\Omega$
- $X_C \approx 31,800,000,000\,\Omega$

It is glaringly obvious that $X_C \gg X_L$ and $X_C \gg R$. The capacitive reactance completely dominates the entire circuit.

### Final Conclusion

Because the capacitor's opposition is so overwhelmingly large compared to the resistor and the inductor, the circuit behaves almost exactly like a pure capacitor.

The phase angle $\phi$ is given by:
$$	an\phi = \frac{X_C - X_L}{R}$$

Since $X_C$ is so huge, $\frac{X_C - X_L}{R}$ approaches infinity. Therefore, the phase angle $\phi$ approaches $90^\circ$.

**The circuit is predominantly capacitive, and the phase angle is approximately $90^\circ$.**

Animated Solution for Physics - Electromagnetic Induction: For a series L-C-R circuit with $R = 100Ω$ , $L = 0.5 mH$ and $C = 0.1 pF$ connected across 220 V-50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is

Select Answer:

Visualized Solution

$L - C - R Circuit Setup$

$Inductive Reactance (X_{L})$

$Substituting Values for X_{L}$

$Calculating X_{L}$

$Capacitive Reactance (X_{C})$

$Substituting Values for X_{C}$

$Calculating X_{C}$

$Comparing Reactances$

$Phase Angle and Circuit Nature$

$The Way Forward$

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Inductive Reactance

The Capacitive Reactance

The Master Comparison

Final Conclusion

Similar Questions

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an AC source $V = V_{0} sin ω t$ as shown. Which of the following options is/are correct?

In a series $L - C - R$ circuit, the inductive reactance ( $X_{L}$ ) is $10 Ω$ and the capacitive reactance ( $X_{C}$ ) is $4 Ω$ . The resistance ( $R$ ) in the circuit is $6 Ω$ . The power factor of the circuit is

In a series resonant $L$ - $C$ - $R$ circuit, the voltage across $R$ is $100 V$ and $R = 1 k Ω$ with $C = 2 μ F$ . The resonant frequency $ω$ is $200 rad/s$ . At resonance, the voltage across $L$ is

In an AC-circuit, an inductor, a capacitor and a resistor are connected in series with $X_{L} = R = X_{C}$ . Impedance of this circuit is

In an $L - C - R$ series AC circuit, the voltage across each of the components $L, C$ and $R$ is $50 V$ . The voltage across the $L - C$ combination will be

An AC circuit has $R = 100Ω$ , $C = 2 μ F$ and $L = 80 mH$ connected in series. The quality factor of the circuit is

In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

List-I

List-II

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L−C−R Circuit Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Inductive Reactance (XL​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for XL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating XL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Capacitive Reactance (XC​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for XC​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating XC​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Reactances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase Angle and Circuit Nature

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward