AC Voltage Applied to Resistor, Inductor & Capacitor: Visual Comparison + Key Derivations

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See how AC voltage behaves across R, L, and C—with intuitive visuals and step-by-step derivations of current, reactance, and phase relationships. Perfect for Class 12, JEE & NEET!

When alternating current (AC) flows through electrical components, each reacts in its own unique way. But why does a resistor “obey” Ohm’s law, while an inductor “fights” change, and a capacitor “stores and releases” energy?

In this visual + derivation-powered guide, we break down what happens when AC voltage is applied to a resistor (R), inductor (L), and capacitor (C)—with clear math you can reproduce in exams and diagrams you’ll remember forever.

Contents hide

1 ⚡ The Setup: Common AC Source

1.1 1. Pure Resistor (R)

1.1.1 🔍 Derivation

1.1.2 📈 Visual Insight

1.2 2. Pure Inductor (L)

1.2.1 🔍 Derivation

1.2.2 📈 Visual Insight

1.3 3. Pure Capacitor (C)

1.3.1 🔍 Derivation

1.3.2 📈 Visual Insight

2 📊 Comparison Table (with Derivation Insights)

3 🎯 Why These Derivations Matter

4 🌟 Final Thought

5 🔗 Dive Deeper on PhysicsQanda.com

⚡ The Setup: Common AC Source

All three circuits are driven by the same sinusoidal voltage:

$v(t) = V_0 \sin(\omega t)$

where:

$V_0$ = peak voltage
$\omega = 2\pi f$ = angular frequency

We’ll derive the instantaneous current $i(t)$ for each case—and reveal the phase difference.

1. Pure Resistor (R)

🔍 Derivation

Ohm’s law holds instantaneously in resistive AC circuits:

$v_R(t) = i_R(t) \cdot R$

Substitute $v_R(t) = V_0 \sin(\omega t)$ :

$i_R(t) = \frac{V_0}{R} \sin(\omega t) = I_0 \sin(\omega t)$

where $I_0 = \frac{V_0}{R}$ .

✅ Phase difference = 0° → voltage and current are in phase.

📈 Visual Insight

Two sine waves—perfectly aligned. No shift.

📌 RMS Form: $V_{\text{rms}} = I_{\text{rms}} R$

2. Pure Inductor (L)

🔍 Derivation

For an inductor, voltage relates to the rate of change of current:

$v_L(t) = L \frac{di_L}{dt}$

Substitute $v_L(t) = V_0 \sin(\omega t)$ :

$L \frac{di_L}{dt} = V_0 \sin(\omega t)$

Integrate both sides:

$i_L(t) = \frac{V_0}{L} \int \sin(\omega t) \, dt = -\frac{V_0}{\omega L} \cos(\omega t)$

Rewrite using $\cos(\theta) = \sin(\theta - \frac{\pi}{2})$ :

$i_L(t) = \frac{V_0}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right) = I_0 \sin\left(\omega t - \frac{\pi}{2}\right)$

where $I_0 = \frac{V_0}{\omega L}$ .

✅ Current lags voltage by 90°.

📈 Visual Insight

Current wave peaks ¼ cycle after voltage.

💡 Define inductive reactance:
$X_L = \omega L = 2\pi f L \quad (\Omega)$

So, $I_0 = \frac{V_0}{X_L}$ —analogous to Ohm’s law!

3. Pure Capacitor (C)

🔍 Derivation

For a capacitor:

$q(t) = C \cdot v_C(t) \quad \text{and} \quad i_C(t) = \frac{dq}{dt}$

So:

$i_C(t) = C \frac{dv_C}{dt} = C \frac{d}{dt}[V_0 \sin(\omega t)] = C V_0 \omega \cos(\omega t)$

Rewrite $\cos(\omega t) = \sin(\omega t + \frac{\pi}{2})$ :

$i_C(t) = \omega C V_0 \sin\left(\omega t + \frac{\pi}{2}\right) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right)$

where $I_0 = V_0 \cdot \omega C$ .

✅ Current leads voltage by 90°.

📈 Visual Insight

Current peaks ¼ cycle before voltage.

💡 Define capacitive reactance:
$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \quad (\Omega)$

Then $I_0 = \frac{V_0}{X_C}$ —again, Ohm’s law form!

📊 Comparison Table (with Derivation Insights)

Component	Current Expression	Phase (I vs V)	Reactance	Key Derivation Step
R	$I_0 \sin(\omega t)$	0°	$R$	$i = v/R$
L	$I_0 \sin(\omega t - \frac{\pi}{2})$	Lags by 90°	$X_L = \omega L$	$v = L \frac{di}{dt}$ → integrate
C	$I_0 \sin(\omega t + \frac{\pi}{2})$	Leads by 90°	$X_C = \frac{1}{\omega C}$	$i = C \frac{dv}{dt}$ → differentiate

🎯 Why These Derivations Matter

NCERT Class 12 Physics (Chapter 7) explicitly asks for derivations of $i(t)$ in L and C.
JEE Main/Advanced often gives differential equation-based problems—mastering these builds intuition.
NEET tests conceptual clarity: “Why does current lead in a capacitor?” is a favorite.

✅ Pro Tip: In exams, always state the governing equation first (e.g., $v = L \frac{di}{dt}$ )—it fetches partial marks even if math falters!

🌟 Final Thought

You now have both visual intuition and mathematical proof—the ultimate combo for mastering AC circuits. Save this guide. Revisit before exams. Teach it to a friend.

🔗 Dive Deeper on PhysicsQanda.com

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AC Voltage Applied to Resistor, Inductor & Capacitor: Visual Comparison + Key Derivations