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 & Intuition

In a capacitor, the relationship between voltage and current is governed by $I = C \frac{dV}{dt}$. When the voltage is sinusoidal, the derivative of sine is cosine, and cosine is just sine shifted forward by 90°. This means the current leads the voltage by a quarter-cycle. No resistance or inductance is present, so the phase shift is exactly ±90° — and because the current must lead, we add π/2 to the sine argument.

Step-by-step reasoning

1. Write the given emf

   The source provides $\varepsilon = \varepsilon_0 \sin(\omega t)$. This is the voltage across the capacitor (since it’s directly connected).

2. Recall the capacitor current–voltage relation

   For a capacitor, $I = C \frac{dV}{dt}$. Here $V = \varepsilon_0 \sin(\omega t)$, so:

3. Express cosine as a shifted sine Using the identity $\cos(\omega t) = \sin(\omega t + \pi/2)$, we get:

4. Define the peak current Let $I_0 = C \varepsilon_0 \omega$. Then:

5. Match with the options This matches option (D) exactly. Option (A) has a minus sign (current lagging), (B) has an extra √2 factor, and (C) has no phase shift — all incorrect for a pure capacitor.

At resonance the capacitive reactance equals the inductive reactance:

With $f=2\text{ MHz}=2\times10^{6}\text{ Hz}$ and $L=20\times10^{-5}\text{ H}=2\times10^{-4}\text{ H}$:

Given: $V = 100\ \mathrm{V}$, $X_L = 24\ \Omega$, $X_C = 16\ \Omega$, $R = 6\ \Omega$, all in series.

Step 1 — Net reactance.

In a series LCR circuit, $V_L$ leads the current by $90^\circ$ and $V_C$ lags it by $90^\circ$ — so they are $180^\circ$ apart and act in opposition. The net reactance is their difference:

Step 2 — Impedance.

The resistive voltage is in phase with the current while the net reactive voltage is $90^\circ$ out of phase, so they add as perpendicular vectors (Pythagoras in the impedance triangle):

(A clean 6–8–10 triangle — a hint that we are on track.)

Step 3 — Current in the circuit.

Series circuit $\Rightarrow$ the same current flows through R, L and C:

Step 4 — Voltage across the series L–C combination.

The individual magnitudes are

But we must not add these arithmetically. Being in antiphase, the resultant across the L–C pair is their difference:

Equivalently, and more directly,

Step 5 — Consistency check.

Across the resistor, $V_R = IR = 10\times 6 = 60\ \mathrm{V}$. Then the total supply voltage should come back out as

which is exactly the source voltage given. The answer is consistent.

Option (B) ($400$ V) comes from adding $V_L + V_C$, the classic error.

ANSWER: A

Concept & Intuition

The bulb’s brightness depends on the power it dissipates, which for a resistive bulb is proportional to the square of the RMS current through it. In a series RC circuit, the total impedance is

As $\omega$ decreases, the capacitive reactance $X_C = 1/(\omega C)$ increases, making $Z$ larger. Since the source voltage amplitude $V_0$ is fixed, the current amplitude $I_0 = V_0 / Z$ decreases. Less current means less power in the resistor — the bulb dims.

Step‑by‑step reasoning

1. Identify the circuit and its impedance The source, capacitor, and resistor are in series. The impedance of a capacitor is purely reactive: $Z_C = 1/(j\omega C)$, so its magnitude is $X_C = 1/(\omega C)$. The total impedance magnitude is

2. Relate current to impedance The source has fixed amplitude $V_0$. The current amplitude is

As $\omega$ decreases, the term $1/(\omega C)$ grows, so the denominator grows — hence $I_0$ decreases.

3. Connect current to bulb brightness The bulb is a pure resistor $R$. The average power dissipated is

A smaller $\omega$ makes the denominator larger, so $P$ becomes smaller. The bulb glows less brightly.

4. Evaluate the options
   • (A) “the bulb switches off” — No, it only dims; current never becomes exactly zero for finite $\omega$.
   • (B) “total impedance is unchanged” — False; impedance increases as $\omega$ decreases.
   • (C) “the bulb glows brighter” — Opposite of what we found.
   • (D) “the bulb glows dimmer” — Correct.

1. Why the resonance condition takes this form

In a series LCR circuit the reactances are

They act in opposition (the inductor voltage leads the current by $90^\circ$, the capacitor voltage lags by $90^\circ$), so the impedance is

At resonance the two cancel, $Z$ falls to its minimum $R$, and the current is maximum. Setting $X_L = X_C$:

And since $\omega_0 = 2\pi f_0$,

2. Substitute the data

3. Finish

4. Watch the trap

Options (C) and (D) drop the $2\pi$ — they quote the angular frequency $\omega_0 = 10^{5}\ \text{rad s}^{-1}$ as though it were the frequency. The question asks for frequency of oscillations in Hz, so the $2\pi$ must stay. Option (B) corresponds to mis-taking $\sqrt{LC}$ as $10^{-1}$.

ANSWER: A

Concept: Series L-C-R — the voltmeter readings are V_R, V_C, V_L across each element (they all carry the same current I), and

V_source = √[V_R² + (V_L − V_C)²].

From the figure: V₁ = 65 V across R, V₂ = 415 V across C, V₃ = 204 V across L.

Consistency check with the source:

√[65² + (415 − 204)²] = √[4225 + 211²] = √[4225 + 44521] = √48746 ≈ 220.8 V ≈ 220 V ✓ — so the assignment of readings is right.

Use the reactance ratio. Since the same current I flows through all three,

X_C/X_L = V_C/V_L = 415/204 = 2.035.

ω = 2π(50) = 314 rad/s.

Test option (B): L = 1.0 H, C = 5 μF

X_L = ωL = 314 × 1.0 = 314 Ω

X_C = 1/(ωC) = 1/(314 × 5 × 10⁻⁶) = 1/(1.57 × 10⁻³) = 637 Ω

Ratio X_C/X_L = 637/314 = 2.03 ✓ — matches 415/204 = 2.035.

Self-consistency: I = V_L/X_L = 204/314 = 0.65 A. Then V_C = I X_C = 0.65 × 637 ≈ 414 V ✓ (given 415 V), and R = V_R/I = 65/0.65 = 100 Ω — all sensible.

Test option (A): L = 2.0 H, C = 5 μF → X_L = 628, X_C = 637, ratio ≈ 1.01 ✗ (needs ≈2.03). (C) and (D) fail similarly.

So L = 1.0 H, C = 5 μF.

ANSWER: B

Concept: in an AC circuit the capacitive reactance is X_C = 1/(omega C), where omega = 2pif.

Hence X_C = 1/(2pif*C).

Dimensional sanity check: with f in Hz (1/s) and C in farads (C/V = As/V), 1/(fC) has units Vs/(A*s) = V/A = ohm, which is correct for a reactance. Note also that X_C falls as f increases (a capacitor blocks DC, f -> 0 gives X_C -> infinity), which only option (D) reproduces.

ANSWER: D

The key idea here is that a transistor amplifier is not an ideal, frequency-independent device. Real amplifiers contain internal and external capacitances — like the coupling capacitors between stages, the bypass capacitor across the emitter resistor, and the transistor's own junction capacitances. These capacitors behave differently at different frequencies.

At low frequencies, the coupling and bypass capacitors have high reactance ($X_C = \frac{1}{2\pi f C}$), so they drop significant signal voltage and reduce the gain. At very high frequencies, the transistor's internal junction capacitances (like $C_{be}$ and $C_{bc}$) start to shunt the signal to ground, again reducing the gain. Only in the middle-frequency range — where these capacitors act as short circuits (for coupling/bypass caps) or open circuits (for junction caps) — does the gain stay flat and maximum.

This gives the amplifier a band-pass characteristic: gain rises from low to mid frequencies, stays constant over a mid-band, then falls off at high frequencies.

Now let's walk through the reasoning step by step.

1. Identify the frequency-dependent components.

   A typical transistor amplifier (common-emitter, for example) has:

   • Coupling capacitors ($C_1$, $C_2$) at input and output — these block DC but pass AC.
   • A bypass capacitor ($C_E$) across the emitter resistor — it shorts the emitter to ground for AC, preventing negative feedback at mid/high frequencies.
   • Transistor internal capacitances ($C_{be}$, $C_{bc}$) — these are always present and become important at high frequencies.

2. Low-frequency behaviour.

   At low $f$, the reactance of $C_1$, $C_2$, and $C_E$ is large.

   • $C_1$ and $C_2$ act as series impedances, dropping voltage and reducing signal transfer.
   • $C_E$ no longer acts as a short; the emitter resistor now provides negative feedback, which reduces the gain. Result: voltage gain is low at low frequencies.

3. Mid-frequency behaviour.

   At mid frequencies, the coupling and bypass capacitors have very low reactance — effectively short circuits. The transistor's junction capacitances still have high reactance — effectively open circuits. So all capacitors are "invisible" to the AC signal, and the amplifier operates at its designed, maximum gain.

   Result: voltage gain is constant and maximum at mid frequencies.

4. High-frequency behaviour.

   At high $f$, the coupling/bypass capacitors are already short circuits (no issue), but the transistor's internal junction capacitances now have low reactance. They shunt the signal from base to emitter or collector to base, reducing the effective input signal and the gain. The transistor's current gain ($\beta$) also falls off at high frequencies.

   Result: voltage gain is low at high frequencies.

5. Combine the three regions.

   The overall gain vs. frequency curve looks like this:

   Frequency range	Voltage gain
   Low	Low (rising)
   Mid	High and constant
   High	Low (falling)

   This matches exactly with option (C).