In an L-R series circuit, a sinusoidal voltage V = V0 sin omega t is applied. It is given that L = 3

Q: Why do we use $V_0$ instead of $V_{\text{rms}}$ to find the amplitude of the current?

The amplitude of the current refers to its peak value ($i_0$). To find the peak current, we must use the peak voltage ($V_0$). Using $V_{\text{rms}}$ would give us the RMS current, not the amplitude.

Q: Why does the current lag the voltage in an L-R circuit?

An inductor opposes changes in current due to self-induction (Lenz's Law). When voltage is applied, the current takes time to build up, causing it to reach its peak after the voltage has already peaked.

Q: How did we get $\omega = 100\pi$?

We are given the frequency $f = \omega / 2\pi = 50 \text{ Hz}$. Multiplying both sides by $2\pi$ gives $\omega = 100\pi \text{ rad/s}$.

Q: Why is the phase shift $\pi/4$ equivalent to a time shift of $T/8$?

A full cycle corresponds to an angle of $2\pi$ and a time of $T$. Therefore, a phase shift of $\pi/4$ corresponds to a time shift of $(\pi/4) / (2\pi) \times T = T/8$.

Q: Can the phase difference in an L-R circuit ever be exactly $\pi/2$?

It can approach $\pi/2$ if the inductive reactance $X_L$ is infinitely larger than the resistance $R$ (a purely inductive circuit). However, as long as there is some resistance, the phase difference will be strictly less than $\pi/2$.

Question

In an $L-R$ series circuit, a sinusoidal voltage $V = V_0 \sin \omega t$ is applied. It is given that $L = 35 	ext{ mH}$, $R = 11 \ \Omega$, $V_{	ext{rms}} = 220 	ext{ V}$, $\omega / 2\pi = 50 	ext{ Hz}$ and $\pi = 22/7$. Find the amplitude of current in the steady state and obtain the phase difference between the current and the voltage. Also plot the variation of current for one cycle on the given graph.

Sigma Percentile · Accepted Answer

The behavior of alternating current in an $L-R$ circuit is a beautiful interplay of resistance and inertia. When a sinusoidal voltage is applied, the inductor resists the change in current, causing the current to lag behind the voltage. Let's break down the mathematics and physics of this phenomenon.

### Analyzing the Setup

We are given an $L-R$ series circuit with the following parameters:
- Inductance, $L = 35 	ext{ mH} = 35 	imes 10^{-3} 	ext{ H}$
- Resistance, $R = 11 \ \Omega$
- RMS Voltage, $V_{	ext{rms}} = 220 	ext{ V}$
- Frequency, $f = \frac{\omega}{2\pi} = 50 	ext{ Hz}$

Our goal is to find the amplitude of the steady-state current ($i_0$), the phase difference ($\phi$), and to visualize the current wave relative to the voltage wave.

### The Master Equation

To find the current amplitude, we first need to determine the total opposition to the current flow, known as the impedance ($Z$). The impedance in an $L-R$ circuit is given by:
$$Z = \sqrt{R^2 + X_L^2}$$
where $X_L$ is the inductive reactance.

Let's calculate $X_L$:
$$X_L = \omega L = 2\pi f L$$
Substituting the given values and using $\pi = \frac{22}{7}$:
$$X_L = 2 	imes \frac{22}{7} 	imes 50 	imes 35 	imes 10^{-3} = 11 \ \Omega$$

Now, we can find the impedance $Z$:
$$Z = \sqrt{11^2 + 11^2} = 11\sqrt{2} \ \Omega$$

### Final Calculation

The amplitude of the current ($i_0$) is the ratio of the peak voltage ($V_0$) to the impedance ($Z$). First, we find the peak voltage from the RMS voltage:
$$V_0 = \sqrt{2} V_{	ext{rms}} = 220\sqrt{2} 	ext{ V}$$

Now, the peak current is:
$$i_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{11\sqrt{2}} = 20 	ext{ A}$$

Next, we determine the phase difference $\phi$. In an $L-R$ circuit, the phase angle is given by:
$$	an \phi = \frac{X_L}{R} = \frac{11}{11} = 1$$
$$\phi = \frac{\pi}{4}$$

Because this is an inductive circuit, the current lags the voltage by $\frac{\pi}{4}$ radians. The equation for the instantaneous current is:
$$i = 20 \sin\left(\omega t - \frac{\pi}{4}ight)$$

Graphically, a phase lag of $\frac{\pi}{4}$ corresponds to a time shift of $\frac{T}{8}$ to the right. The current wave crosses zero at $t = \frac{T}{8}$ and reaches its peak at $t = \frac{3T}{8}$, perfectly illustrating the "inertia" of the inductor!

Visualized Solution

$V = V_{0} sin ω t$

$X_{L} and Z$

$Calculating Z$

$i_{0} = \frac{V _{0}}{Z}$

$tan ϕ = \frac{X _{L}}{R}$

$i = i_{0} sin (ω t - ϕ)$

\text{Conclusion}

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=V0​sinωt

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i0​=ZV0​​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i=i0​sin(ωt−ϕ)

\text{Conclusion}

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Final Calculation

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$Calculating Z$

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