Q.A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?

Concept and Intuition

In a series R-C circuit, $X_C=\dfrac{1}{2\pi fC}$ is inversely proportional to frequency, while $R$ stays fixed. The impedance $Z=\sqrt{R^2+X_C^2}$ sets the current $I=V/Z$ at constant voltage. Lowering the frequency raises $X_C$ (capacitor becomes more 'blocking'), raising $Z$ and lowering $I$ — exactly the behaviour described.

Step-by-Step Solution

1. Let the original frequency be $f$, with reactance $X_C$ and impedance $Z_1=\sqrt{R^2+X_C^2}$, current $I=V/Z_1$.
2. At frequency $f/3$: $X_C'=\dfrac{1}{2\pi(f/3)C}=3X_C$. New impedance $Z_2=\sqrt{R^2+9X_C^2}$.
3. Given new current is $I/2$, and $V$ unchanged: $Z_2=2Z_1$, so $Z_2^2=4Z_1^2$.
4. $R^2+9X_C^2=4(R^2+X_C^2)=4R^2+4X_C^2$.
5. $9X_C^2-4X_C^2=4R^2-R^2\Rightarrow5X_C^2=3R^2\Rightarrow\left(\dfrac{X_C}{R}\right)^2=\dfrac{3}{5}\Rightarrow\dfrac{X_C}{R}=\left(\dfrac{3}{5}\right)^{1/2}$.

Common Mistakes

• Forgetting that halving the frequency scales $X_C$ by 3 (not by 1/3) since $X_C\propto1/f$.
• Squaring the impedance ratio incorrectly (using $Z_2=2Z_1$ directly instead of $Z_2^2=4Z_1^2$ inside the Pythagorean sum).

ANSWER: A

Concept

In a single series loop the current is the same everywhere, so the ammeter $A$ measures the current common to $C$ and $R$. The voltmeter $V_1$ is across the capacitor and $V_2$ is across the resistor. The two fixed facts you need:

• Across a resistor, voltage and current are in phase.
• Across a capacitor, the current leads the voltage by $90^\circ$ (ELI the ICE-man: in a Capacitor, I leads E).

Statement I - $A$ and $V_2$ are in phase

$V_2$ is the resistor voltage, which is always in phase with the current read by $A$. True.

Statement II - $V_1$ leads $V_2$

$V_2$ (resistor) is in phase with the current, while $V_1$ (capacitor) lags the current by $90^\circ$. Hence $V_1$ lags $V_2$ by $90^\circ$; it does not lead it. False.

Statement III - $A$ leads $V_1$

$A$ is the current and $V_1$ is the capacitor voltage. Since current leads capacitor voltage by $90^\circ$, $A$ leads $V_1$. True.

Only I and III hold.

Concept and Intuition

A capacitor's opposition to alternating current, the capacitive reactance, is $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$. It decreases with increasing frequency because a capacitor charges and discharges more easily (offers less opposition) when the applied voltage reverses direction faster.

Step-by-Step Solution

1. Given: $C = 500\ \text{nF} = 500\times10^{-9}\ \text{F} = 5\times10^{-7}\ \text{F}$, $f = 1\ \text{kHz} = 10^3\ \text{Hz}$.
2. $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 10^3 \times 5\times10^{-7}}$
3. Compute the denominator: $2\pi\times10^3\times5\times10^{-7} = 2\pi \times 5\times10^{-4} = 10\pi\times10^{-4} = \pi\times10^{-3}$.
4. So $X_C = \frac{1}{\pi\times10^{-3}} = \frac{10^3}{\pi}\ \Omega$

Common Mistakes

• Forgetting the factor of $2\pi$ (using $f$ instead of $\omega=2\pi f$).
• Unit slip when converting nF to F (500 nF = $5\times10^{-7}$ F, not $5\times10^{-9}$ F times 500 loosely miscomputed).

ANSWER: A

Concept and Intuition

A capacitor in an AC circuit offers an opposition to current flow called capacitive reactance, $X_C = \dfrac{1}{\omega C}$. Unlike a resistor, this reactance depends inversely on capacitance: a bigger capacitor 'blocks' AC less. Since the bulb (a resistor) is in series with the capacitor, the same current flows through both. The total impedance of the series combination is

Step-by-Step Solution

1. Capacitive reactance: $X_C=\dfrac{1}{\omega C}$. As $C$ increases, $X_C$ decreases.
2. Impedance of the series R–C circuit: $Z=\sqrt{R^2+X_C^2}$. Since $X_C$ decreases, $Z$ decreases.
3. Current in the circuit: $I=\dfrac{V_{rms}}{Z}$. Since $Z$ decreases while $V_{rms}$ (supply) is fixed, $I$ increases.
4. Power dissipated in the bulb: $P=I^2R$ (only the resistor dissipates real power; the capacitor stores/releases energy with zero average dissipation). Since $I$ increases, $P$ increases.
5. Hence the bulb glows brighter — its power output increases.

Common Mistakes

• Thinking capacitors 'consume' power like resistors — they don't; only $R$ dissipates real power, but a larger current through $R$ still raises $P$.
• Confusing capacitive reactance's inverse dependence on $C$ with a direct one, and concluding the opposite trend.

ANSWER: B

Concept and Intuition

In a series RC AC circuit, impedance is $Z=\sqrt{R^2+X_C^2}$ where $X_C=\dfrac{1}{\omega C}$ is the capacitive reactance. Lowering the frequency raises $X_C$ (harder for the capacitor to "pass" current), which raises $Z$ and lowers the current for the same applied voltage — matching the given fact that current drops when frequency drops. The power factor $\cos\phi=R/Z$ measures how resistive (vs. reactive) the circuit's response is at a given frequency; it needs both $R$ and $X_C$ at that specific frequency.

Step-by-Step Solution

1. Let the initial angular frequency be $\omega_0$, reactance $X_0=1/(\omega_0 C)$, impedance $Z_0=\sqrt{R^2+X_0^2}$, current $I_0=V/Z_0$.
2. At $\omega_0/3$: reactance becomes $X'=1/((\omega_0/3)C)=3X_0$ (reactance is inversely proportional to frequency).
3. Current halves: $I'=I_0/2=V/Z' \Rightarrow Z'=2Z_0$.
4. Square both impedances: $Z_0^2=R^2+X_0^2$ and $Z'^2=R^2+(3X_0)^2=R^2+9X_0^2$.
5. Substitute $Z'^2=4Z_0^2$: $R^2+9X_0^2=4(R^2+X_0^2)=4R^2+4X_0^2$.
6. Rearranging: $5X_0^2=3R^2 \Rightarrow X_0^2=\dfrac{3}{5}R^2$.
7. Power factor at the initial frequency: $\cos\phi=\dfrac{R}{Z_0}=\dfrac{R}{\sqrt{R^2+\frac35R^2}}=\dfrac{R}{R\sqrt{8/5}}=\sqrt{\dfrac{5}{8}}$.

Common Mistakes

• Forgetting that reactance is inversely proportional to frequency, and instead scaling it the wrong way (multiplying rather than dividing, or vice versa).
• Squaring the "current halves" condition incorrectly — remember $Z\propto 1/I$ for fixed voltage, so halving current means doubling, not halving, the impedance.

ANSWER: C

Concept and Intuition

A capacitor's opposition to AC current, $X_C=\dfrac{1}{2\pi fC}$, decreases as frequency increases — at higher frequency the capacitor charges and discharges faster, letting more current through for the same voltage.

Step-by-Step Solution

1. $X_{C,1}=\dfrac{1}{2\pi f C}=3\,\mathrm{k}\Omega$ at frequency $f$.
2. At frequency $2f$: $X_{C,2}=\dfrac{1}{2\pi(2f)C}=\dfrac{X_{C,1}}{2}$.
3. $X_{C,2}=\dfrac{3}{2}=1.5\,\mathrm{k}\Omega$.

Common Mistakes

• Confusing capacitive reactance's inverse relation with frequency with inductive reactance's direct proportionality ($X_L\propto f$), which would (wrongly) double it.

ANSWER: A

Concept and Intuition

A capacitor's opposition to AC is the reactance $X_C = 1/(\omega C)$, which plays the role of "resistance" in $V=IX_C$ for rms quantities in a purely capacitive AC circuit.

Step-by-Step Solution

1. $\omega = 2\pi f = 2\pi(50) = 100\pi$ rad/s.
2. $X_C = \dfrac{1}{\omega C} = \dfrac{1}{100\pi\times\left(\dfrac{50}{\pi}\times10^{-6}\right)}$.
3. The $\pi$'s cancel: $100\pi\times\dfrac{50}{\pi}\times10^{-6} = 100\times50\times10^{-6} = 5000\times10^{-6} = 5\times10^{-3}$.
4. $X_C = \dfrac{1}{5\times10^{-3}} = 200\ \Omega$.
5. $I_{rms} = \dfrac{V_{rms}}{X_C} = \dfrac{250}{200} = 1.25$ A.

Common Mistakes

• Forgetting to convert $f$ to $\omega$ (using $f$ directly in $X_C=1/(fC)$).
• Arithmetic slip in cancelling the $\pi$ factor between $\omega$ and $C$.

ANSWER: A

Concept and Intuition

In an AC circuit, a capacitor opposes current with a frequency-dependent reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ (unlike a resistor, this depends on frequency). Once $X_C$ is known, rms current follows exactly as in a resistive circuit: $I_{rms}=V_{rms}/X_C$.

Step-by-Step Solution

1. $C = 100\ \mu\text{F} = 10^{-4}\ \text{F}$, $f=50$ Hz, $V_{rms}=100$ V.
2. $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi(50)(10^{-4})} = \dfrac{1}{0.0314159} \approx 31.83\ \Omega$.
3. $I_{rms} = \dfrac{V_{rms}}{X_C} = \dfrac{100}{31.83} \approx 3.14\ \text{A}$.

Common Mistakes

• Using $X_C = \omega C$ instead of $1/(\omega C)$.
• Forgetting to convert $\mu\text{F}$ to F before substituting.

ANSWER: A

Concept and Intuition

Capacitive reactance $X_C = \dfrac{1}{2\pi f C}$ decreases as frequency increases — a capacitor "conducts" AC more easily at higher frequencies. Since $C$ is fixed here, $X_C \propto 1/f$ directly.

Step-by-Step Solution

1. Original reactance: $X_C = \dfrac{1}{2\pi f C} = 6\,\mathrm{k}\Omega$.
2. New frequency: $f' = 2f$. New reactance: $X_C' = \dfrac{1}{2\pi(2f)C} = \dfrac{X_C}{2}$.
3. $X_C' = \dfrac{6}{2} = 3\,\mathrm{k}\Omega$.

Common Mistakes

• Assuming reactance is proportional to frequency (that is true for inductive reactance $X_L=\omega L$, not capacitive reactance).

ANSWER: B

Concept and Intuition

The current through a capacitor is $I=C\dfrac{dV}{dt}$. If $V=V_0\sin\omega t$, then $I=C V_0\omega\cos\omega t = I_0\sin(\omega t+90^\circ)$ — the current waveform is a cosine relative to the voltage sine, i.e., it peaks a quarter cycle before the voltage.

Step-by-Step Solution

1. Let applied voltage $V=V_0\sin\omega t$.
2. Current through capacitor: $I=C\dfrac{dV}{dt}=C V_0\omega\cos\omega t$.
3. Rewrite $\cos\omega t=\sin(\omega t+90^\circ)$, so $I=I_0\sin(\omega t+90^\circ)$ where $I_0=\omega C V_0$.
4. Comparing phases, current is ahead of (leads) voltage by $90^\circ$.

Common Mistakes

• Confusing capacitive circuits (current leads) with inductive circuits (current lags) — a common mnemonic is "CIVIL": in a Capacitor, I leads V; in an inductor, V leads I (L).
• Thinking the phase difference is $180^\circ$ instead of $90^\circ$.

ANSWER: C

Concept and Intuition

For a series R-C circuit, impedance is $Z=\sqrt{R^2+X_C^2}$ with $X_C = \dfrac{1}{\omega C}$. Lowering the frequency increases $X_C$ (reactance rises as frequency falls), which increases impedance and hence decreases current for the same applied voltage. Given how much the current drops when frequency is reduced to $\omega/3$, we can back out the reactance-to-resistance ratio at the original frequency.

Step-by-Step Solution

1. At frequency $\omega$: current $I = V/Z_1$, where $Z_1=\sqrt{R^2+X_C^2}$ and $X_C=1/(\omega C)$.
2. At frequency $\omega/3$: reactance becomes $X_C' = \dfrac{1}{(\omega/3)C} = 3X_C$. Current becomes $I/2 = V/Z_2$, where $Z_2=\sqrt{R^2+9X_C^2}$.
3. Since voltage $V$ is the same in both cases:

4. Square both sides:

5. Rearrange:

6. Solve for the ratio:

Common Mistakes

• Forgetting that lowering frequency to $\omega/3$ triples (not divides) the capacitive reactance, since $X_C\propto1/\omega$.
• Sign/algebra slip when expanding $(2Z_1)^2$ or rearranging terms.

ANSWER: C