Q.A resistor of $200\ \Omega$ and a capacitor of $15.0\ \mu\text{F}$ are connected in series to a $220\ \text{V}$, $50\ \text{Hz}$ ac source.

Concept: Capacitive Reactance — in an AC circuit, the resistor's voltage is in phase with the current while the capacitor's voltage lags the current by $90^\circ$, so their rms values do not add algebraically.

(a)

Capacitive reactance:

$X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \times 50 \times 15.0 \times 10^{-6}} \approx 212.2\ \Omega$

Impedance:

$Z = \sqrt{R^2 + X_C^2} = \sqrt{200^2 + 212.2^2} \approx 291.5\ \Omega$

Current (rms):

$I = \dfrac{V}{Z} = \dfrac{220}{291.5} \approx 0.755\ \text{A}$

(b)

Voltage across resistor: $V_R = I R = 0.755 \times 200 \approx 151\ \text{V}$

Voltage across capacitor: $V_C = I X_C = 0.755 \times 212.2 \approx 160\ \text{V}$

Algebraic sum: $151 + 160 = 311\ \text{V}$, which is greater than $220\ \text{V}$. …

Concept and Intuition

When a resistor and capacitor are in series with an AC source, the resistor's voltage is in phase with the current, while the capacitor's voltage lags the current by $90^\circ$. This phase difference means the two voltages do not peak at the same time — so you cannot simply add their RMS values arithmetically. Instead, the total voltage is the phasor sum, which is the hypotenuse of a right triangle: $V = \sqrt{V_R^2 + V_C^2}$.

The impedance $Z$ of the series RC circuit is the AC analogue of resistance: $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac{1}{2\pi f C}$ is the capacitive reactance. Ohm's law for AC gives $I = V/Z$.

Step-by-Step Solution

1. Find the capacitive reactance $X_C$.

The formula is:

Given $f = 50\ \text{Hz}$ and $C = 15.0\ \mu\text{F} = 15.0 \times 10^{-6}\ \text{F}$:

2. Compute the total impedance $Z$ of the series circuit.

Since $R$ and $X_C$ are orthogonal (resistor voltage in phase, capacitor voltage $90^\circ$ behind), impedance adds like the sides of a right triangle:

3. Calculate the RMS current $I$ in the circuit.

Using Ohm's law for AC:

So the current is about 0.755 A.

4. Find the RMS voltage across the resistor, $V_R$.

The resistor obeys Ohm's law with no phase shift:

5. Find the RMS voltage across the capacitor, $V_C$.

Similarly:

6. Check the algebraic sum and resolve the paradox. …

Method: Phasor Analysis of Series RC Circuit

This method uses phasor addition (not algebraic addition) because voltage and current are out of phase in reactive circuits.

Step-by-step solution

Step 1: Calculate capacitive reactance

Step 2: Calculate impedance of the series circuit

Step 3: Calculate current (rms) — part (a)

Step 4: Calculate voltage across resistor and capacitor — part (b)

• Across resistor: $V_R = I \times R = 0.754 \times 200$

• Across capacitor: $V_C = I \times X_C = 0.754 \times 212.2$

Step 5: Check the algebraic sum

Yes, the algebraic sum exceeds the source voltage.

Step 6: Resolve the paradox …

Common Mistakes & How to Avoid Them

Mistake 1: Using DC formulas directly for AC circuits

The error: Students often calculate current as $I = \frac{V}{R} = \frac{220}{200} = 1.1\ \text{A}$, ignoring the capacitor entirely.

Why it's wrong: In an AC circuit with a capacitor, the opposition to current is impedance ($Z$), not just resistance. The capacitor adds capacitive reactance ($X_C$).

How to avoid: Always identify all components. For series R-C circuits, use:

where $X_C = \frac{1}{2\pi f C}$.

Mistake 2: Forgetting to convert units properly

The error: Using $C = 15.0$ instead of $C = 15.0 \times 10^{-6}\ \text{F}$.

Why it's wrong: $1\ \mu\text{F} = 10^{-6}\ \text{F}$. Using $15$ gives $X_C = \frac{1}{2\pi (50)(15)} \approx 2.12 \times 10^{-4}\ \Omega$, which is absurdly small.

How to avoid: Write all quantities in SI units before calculation:

• $C = 15.0\ \mu\text{F} = 15.0 \times 10^{-6}\ \text{F}$
• $f = 50\ \text{Hz}$
• $R = 200\ \Omega$

Mistake 3: Adding voltages arithmetically without considering phase

The error: Computing $V_R + V_C = IR + IX_C$ and comparing directly with source voltage $V$.

Why it's wrong: In AC circuits, voltages across R and C are out of phase by $90^\circ$. The resistor voltage ($V_R$) is in phase with current, while capacitor voltage ($V_C$) lags current by $90^\circ$. They add as vectors, not scalars.

How to avoid: Use phasor addition:

This is always less than $V_R + V_C$ (unless one is zero).

Mistake 4: Confusing RMS, peak, and instantaneous values

The error: Using $V = 220\ \text{V}$ as peak voltage, or mixing RMS and peak in the same formula.

Why it's wrong: The given $220\ \text{V}$ is RMS (standard for AC mains). Ohm's law for AC uses RMS values consistently:

How to avoid: Always check:

• Source voltage given → assume RMS unless stated otherwise
• All calculations use RMS values
• If peak is needed: $V_0 = V_{\text{rms}} \times \sqrt{2}$

Mistake 5: Not resolving the "paradox" correctly

The error: Saying "the algebraic sum is more because of some error" or "it's impossible." …