In a series resonant L-C-R circuit, the voltage across R is 100 V and R = 1 kΩ with C = 2 µF. The re

Q: Why is the current calculated using only the resistor's voltage?

Because Ohm's law $V_R = I R$ applies locally to the resistor at all times. Since we are given both $V_R$ and $R$, it's the most direct way to find the series current.

Q: Why didn't we find the value of $L$ first?

We absolutely could! You could use $\omega L = \frac{1}{\omega C}$ to find $L$, and then plug it into $V_L = I \omega L$. However, substituting the expression directly saves a step and reduces the chance of calculation errors.

Q: Is the voltage across the inductor and capacitor always equal at resonance?

Yes! At resonance, $X_L = X_C$. Since the same current $I$ flows through both, their voltage magnitudes $V_L = I X_L$ and $V_C = I X_C$ are perfectly equal. However, they are $180^\circ$ out of phase, so they cancel each other out perfectly.

Q: Why is the total source voltage not given?

We simply don't need it! The local voltage across the resistor gives us the current. Interestingly, at resonance, the total source voltage is actually exactly equal to the voltage across the resistor, so the source voltage is also $100\text{ V}$.

Q: What does 'series resonant' mean physically?

Physically, it means the circuit is driven at a 'Goldilocks' frequency where the energy sloshes perfectly back and forth between the inductor's magnetic field and the capacitor's electric field. They trade energy so efficiently that the source only needs to supply energy to overcome the resistor's heat loss.

Question

In a series resonant $L$-$C$-$R$ circuit, the voltage across $R$ is $100	ext{ V}$ and $R =1	ext{ k}\Omega$ with $C =2\mu	ext{F}$. The resonant frequency $\omega$ is $200	ext{ rad/s}$. At resonance, the voltage across $L$ is

Options:

(A) $2.5 	imes 10^{-2}	ext{ V}$

(B) $40	ext{ V}$

(C) $250	ext{ V}$

(D) $4 	imes 10^{-3}	ext{ V}$

Sigma Percentile · Accepted Answer

### The Anatomy of a Resonant Circuit

Imagine a playground swing. If you push it at just the right rhythm, it swings higher and higher with very little effort. This is the essence of **resonance**. In the world of electronics, an $L$-$C$-$R$ series circuit behaves exactly like that swing when driven by an AC source at its resonant frequency.

In our specific problem, we are given a series $L$-$C$-$R$ circuit. The voltage across the resistor is $V_R = 100	ext{ V}$, and its resistance is $R = 1	ext{ k}\Omega = 1000	ext{ }\Omega$. The capacitance is $C = 2\mu	ext{F} = 2 	imes 10^{-6}	ext{ F}$, and the circuit is driven at its resonant frequency, $\omega = 200	ext{ rad/s}$. Our goal is to find the voltage across the inductor, $V_L$.

### The Master Key: The Resonance Condition

The most critical piece of information given is that the circuit is at **resonance**. What does this mean physically? It means the inductor and the capacitor are trading energy back and forth perfectly. Mathematically, it means their reactances are exactly equal:

$$X_L = X_C$$

Since we know that $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$, we can write our master equation:

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

This relationship is the golden key that will unlock the rest of the problem, allowing us to bypass the missing value of the inductance $L$.

### Finding the Flow: Circuit Current

Before we can find the voltage across the inductor, we need to know how much current is flowing through it. Because this is a series circuit, the current $I$ is identical through every single component—the resistor, the inductor, and the capacitor.

We have the voltage across the resistor and its resistance. Ohm's law applies locally to the resistor at all times, so we can easily find the current:

$$I = \frac{V_R}{R}$$

Substituting our known values:

$$I = \frac{100	ext{ V}}{1000	ext{ }\Omega} = 0.1	ext{ A}$$

So, a steady RMS current of $0.1	ext{ A}$ is flowing through the entire circuit.

### The Inductor's Secret

Now, we turn our attention to the inductor. The voltage across the inductor is given by the product of the current and its inductive reactance:

$$V_L = I X_L = I \omega L$$

Here is where many students get stuck. We don't know $L$! But remember our master key? We established that at resonance, $\omega L = \frac{1}{\omega C}$. We can substitute this entire expression directly into our voltage equation:

$$V_L = I \left(\frac{1}{\omega C}ight) = \frac{I}{\omega C}$$

This elegant substitution saves us from having to calculate $L$ separately and keeps our math clean.

### The Final Calculation

All that remains is to plug in the numbers and carefully execute the arithmetic.

$$V_L = \frac{0.1}{200 	imes 2 	imes 10^{-6}}$$

First, let's simplify the denominator:

$$200 	imes 2 	imes 10^{-6} = 400 	imes 10^{-6} = 4 	imes 10^{-4}$$

Now, substitute this back into the fraction:

$$V_L = \frac{0.1}{4 	imes 10^{-4}}$$

To make this easier to compute, we can multiply the numerator and denominator by $10^4$:

$$V_L = \frac{0.1 	imes 10^4}{4} = \frac{1000}{4} = 250	ext{ V}$$

And there we have it! The voltage across the inductor is **$250	ext{ V}$**.

As a fascinating side note, because the circuit is at resonance, the voltage across the capacitor $V_C$ is also exactly $250	ext{ V}$. However, because they are $180^\circ$ out of phase, they perfectly cancel each other out, leaving the total source voltage to be exactly equal to the resistor's voltage: $100	ext{ V}$.

Animated Solution for Physics - Electromagnetic Induction: 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

Select Answer:

Visualized Solution

$L$ - $C$ - $R$ Circuit Setup

Resonance Condition

Circuit Current

Calculating Current

Voltage across Inductor

Substituting $L$

Final Calculation

Final Answer

Conceptual Takeaway

The Sigma Insight: Alternating Current (AC) and Voltage

The Anatomy of a Resonant Circuit

The Master Key

Finding the Flow

The Inductor's Secret

The Final Calculation

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