Question

(a) In an RLC circuit, can the amplitude of the voltage across an inductor be greater than the amplitude of the generator emf? (b) Consider an RLC circuit with emf amplitude ${\mathbf{\in }}_{\mathbf{m}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\text{V}}$, resistance$\mathit{R}\mathbf{=}\mathbf{10}\mathbf{\text{Ω}}$ , inductance$\mathit{L}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{H}}$ , and capacitance$\mathit{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{μF}}$ . Find the amplitude of the voltage across the inductor at resonance.

Short answer

1. Yes; the amplitude of the voltage across an inductor can be greater than the generator emf.
2. Amplitude of voltage across inductor at resonance is $1.0\times {10}^3\text{V}$.

Step-by-step solution

The given data

1. Resistance,$R=10\text{Ω}$
2. Inductance,$L=1.0\text{H}$
3. Capacitance, role="math" localid="1662974163991" $C=1.0\text{μF}$
4. Amplitude of emf,role="math" localid="1662974170414" ${\in }_{\mathrm{m}}=10\text{V}$

Understanding the concept of reactance and impedance

At resonance, the driving angular frequency is equal to the natural angular frequency of the circuit. Using this concept, we can find inductive reactance. At resonance, the capacitive reactance has the same value as the inductive reactance.

Formulae:

1. Condition for resonance for series $LCR$circuit $X_C=X_L$and $\varphi =0$ ...(1)
2. Inductive reactance, $X_L={\omega }_{d}L$ ...(2)
3. Current amplitude using Ohm’s law, $I=\frac{{\epsilon }_m}{Z}$ ...(3)
4. The resonance frequency of LC oscillations, $\omega =\frac{1}{\sqrt{LC}}$ ...(4)

Here Lis the inductance of the inductor, C is the capacitance of the capacitor and ${\omega }_d$ is the driving angular frequency.

a) Calculation of the voltage amplitude

The amplitude of the voltage across an inductor can be greater than the amplitude of the generator emf in an RLC circuit.

b) Calculation of voltage amplitude across the inductor

At resonance, driving angular frequency ${\omega }_d$is equal to the natural angular frequency$\omega$.

Thus, the inductive reactance using equation (4) in equation (2) as follows:

$$\begin{gathered} X_L=\frac{L}{\sqrt{LC}} \\ =\frac{1.0\text{H}}{\sqrt{\left(1.0\text{H}\right)\left(1.0\times {10}^{-6} \text{F}\right)}} \\ =1000\text{Ω} \end{gathered}$$

At resonance, the capacitive reactance has the same value as the inductive reactance, i.e., given using equation (i)

So the impedance reduces to$Z=R$.

Now, the current amplitude is given using equation (3) as follows:

$$\begin{gathered} I=\frac{{\epsilon }_m}{R} \\ =\frac{10\text{V}}{10\text{Ω}} \\ =1.0\text{A} \end{gathered}$$

Voltage amplitude across the inductor in RLC circuit is given using equation (3) as follows:

$$\begin{gathered} V_L=\left(1.0\text{A}\right)\left(1000\text{Ω}\right) \\ =1.0\times {10}^3\text{V} \end{gathered}$$

This is much greater than the amplitude of the generator emf.

Hence, the value of the amplitude voltage is $1.0\times {10}^3\text{V}$.