Q.Figure 7.17 shows a series LCR circuit connected to a variable frequency $230\ \text{V}$ source. $L = 5.0\ \text{H}$, $C = 80\ \mu\text{F}$, $R = 40\ \Omega$.

Step 1 — What the mains actually delivers.

Electricity is generated by AC generators (alternators) and distributed to homes as an alternating voltage. In India it is

so the peak value is $V_0 = \sqrt{2}\times 230 \approx 325\ \mathrm{V}$, and the polarity reverses $2f = 100$ times each second.

Step 2 — Why the current must also be AC.

The current in any element of the household circuit is determined by that applied voltage and the element's impedance $Z$:

Because $V$ oscillates sinusoidally and reverses sign every half-cycle, so does $I$. There is no passive element that can convert an alternating driving voltage into a steady one-directional current — that requires an active rectifier (a diode circuit), which is exactly what appliances that need DC (chargers, LED drivers, TVs) contain inside themselves. The mains supply itself, before rectification, is AC through and through.

Step 3 — Why AC is used at all. …

Concept & Intuition

In alternating current (AC) circuits, the phase relationship between voltage and current depends on the circuit’s impedance.

• A resistor has no phase shift: voltage and current rise and fall together.
• An inductor causes current to lag voltage by 90°.
• A capacitor causes current to lead voltage by 90°. When these reactive elements are combined, the net phase shift is determined by the difference between inductive and capacitive reactance. Only when the total reactance is zero (purely resistive) do voltage and current align perfectly in phase.

Step-by-step reasoning

1. Purely resistive circuit (Option A)

   • Impedance is just $Z = R$ (real, no imaginary part).
   • Ohm’s law: $V = IR$ holds instantaneously.
   • Therefore, $V$ and $I$ are exactly in phase.
   • ✓ This is the condition we are looking for.

2. Series LC circuit (Option B)

   • Impedance $Z = j(\omega L - \frac{1}{\omega C})$, purely imaginary.
   • At resonance, $\omega L = \frac{1}{\omega C}$, so $Z = 0$ (theoretically a short circuit).
   • Even at resonance, the voltage and current are not in phase — the circuit behaves as a pure resistance only if there is also a resistor. Without resistance, the phase is either +90° or –90° (or undefined at resonance).
   • ✗ Not in phase.

3. Purely inductive circuit (Option C)

   • Impedance $Z = j\omega L$, purely imaginary positive.
   • Current lags voltage by exactly $90^\circ$.
   • ✗ Not in phase.

4. Purely capacitive circuit (Option D) …

Concept and intuition:

In an LCR series circuit, the average power transferred over a complete cycle depends on the phase angle between voltage and current. At the natural (resonant) frequency, the inductive and capacitive reactances cancel each other ($X_L = X_C$), so the impedance is purely resistive and minimum ($Z = R$). The power factor becomes 1, meaning all the power supplied by the source is dissipated in the resistor. Therefore, the average power is simply the DC power formula $P = V_{\text{rms}}^2 / R$.

Step-by-step solution:

1. Identify the condition: The problem states the supply frequency equals the natural (resonant) frequency of the circuit. At resonance, $X_L = X_C$, so the total impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = R$.

2. Recall the average power formula: For an AC circuit, the average power over one cycle is

where $\cos\phi$ is the power factor. At resonance, $\phi = 0$, so $\cos\phi = 1$.

3. Find the rms current: Since $Z = R$, the rms current is

4. Substitute into power formula: $P_{\text{avg}} = V_{\text{rms}} \cdot \frac{V_{\text{rms}}}{R} \cdot 1 = \frac{V_{\text{rms}}^2}{R}.$ …

Concept. In a series LCR circuit at resonance, $X_L=X_C$, so the reactive voltages across L and C cancel and the impedance is purely resistive, $Z=R$. The whole supply voltage then drops across $R$.

Step 1 — Initial conditions.

Given $V_R=V_L=V_C=10\,$V at resonance. Since $Z=R$, the supply voltage is

The reactances (fixed L and C) satisfy $I_0 X_L = I_0 X_C = 10\,$V.

Step 2 — Halve the resistance.

The source voltage is unchanged, so with $R' = R/2$ the new current is

Step 3 — New voltage across $R$. …

For a series LCR circuit the supply voltage relates to the element voltages by

Given $V_1=V_L=100$ V and $V_2=V_C=100$ V, so $V_L-V_C=0$.

Hence $V_R=V_3=\sqrt{V^2-0}=V=220$ V. …

Angular frequency:

Reactances:

$X_C=\frac{1}{\omega C}=\frac{1}{100\times 20\times10^{-6}}=\frac{1}{2\times10^{-3}}=500\,\Omega.$ …

With $L$ removed, the R–C circuit has $\tan\phi=\dfrac{X_C}{R}=\tan\dfrac{\pi}{3}$, so $X_C=\sqrt3\,R$.

With $C$ removed, the R–L circuit has $\tan\phi=\dfrac{X_L}{R}=\tan\dfrac{\pi}{3}$, so $X_L=\sqrt3\,R$. …

Concept: series LCR resonance occurs at f = 1/(2pisqrt(LC)), so at a fixed resonant frequency the product LC must stay constant.

L1 C1 = L2 C2

L * C = (L/3) * C2

=> C2 = 3C. …

In a series LCR circuit, power is maximum at the resonant frequency, where $X_L = X_C$ and the impedance equals $R$. Then

$P_{max} = \frac{V_{rms}^2}{R} = \frac{(230)^2}{23} = \frac{52900}{23} = 2300\ \text{W}.$ …

The key idea here is that an antenna works efficiently only when its physical size is of the same order as the wavelength of the electromagnetic wave it is meant to transmit or receive. This is not an arbitrary rule — it comes from the physics of how oscillating charges radiate energy. If the antenna is too small compared to the wavelength, the radiated power is negligible; if it is much larger, it becomes impractical and directional.

For a simple dipole or monopole antenna, the most common design is a quarter-wavelength ($\lambda/4$) antenna. This length gives a good impedance match and efficient radiation. The problem asks for the "proper working" size, which in standard exam context means the minimum practical size for efficient radiation — and that is $\lambda/4$.

Let’s work it out.

1. Find the wavelength of the electromagnetic wave. The relationship between frequency $f$ and wavelength $\lambda$ for any EM wave in vacuum (or air) is:

where $c = 3 \times 10^8$ m/s is the speed of light.

2. Plug in the given frequency $f = 5$ MHz $= 5 \times 10^6$ Hz:

3. Determine the antenna size for proper working. The standard rule: the antenna length $L$ should be at least $\lambda/4$ for efficient radiation. So: $L = \frac{\lambda}{4} = \frac{60}{4} = 15 \text{ m}$ …

Step 1 — Establish that every resistor sees the full 300 V

All five resistors hang between the same two rails, and those rails are connected directly to the two battery terminals. Since an ideal ammeter has zero resistance, it drops no voltage and does not disturb the potentials of the rails. Therefore each resistor has the full 300 V across it, independent of the others — that is the defining property of a parallel combination.

Step 2 — Current in each branch (Ohm's law)

This is the same for all five: $I_1 = I_2 = I_3 = I_4 = I_5 = 0.2\ \text{A}$.

(For reference, the total current drawn from the battery is $5 \times 0.2 = 1$ A, consistent with $R_{\text{eq}} = 1500/5 = 300\,\Omega$ and $I = 300/300 = 1$ A.)

Step 3 — Decide which currents actually pass through the ammeter

This is the whole point of the question — an ammeter reads only the current in the wire segment it is inserted into.

The ammeter lies in the bottom rail, between the bottom node of R2 and the bottom node of R3. Trace the return paths back to the battery's negative terminal (which is on the left):

• R1 and R2 are already to the left of the ammeter. Their return currents reach the battery without crossing it. ⇒ not counted.
• R3, R4 and R5 are to the right of the ammeter. Their return currents must travel leftwards along the bottom rail through the ammeter to get back to the battery. ⇒ counted. …

Step 1 — Why rms and not peak.

An AC voltage varies as $v = V_0\sin\omega t$. Its average over a cycle is zero, so it is useless as a rating. The meaningful measure is the root-mean-square value, defined so that it delivers the same heating power as a DC source of the same value:

All AC voltmeters and all supply specifications are calibrated in rms. So $220\ \mathrm{V}$ is $V_{rms}$; the peak value is actually

which is not what is quoted — ruling out (A) and (D).

Step 2 — Frequency vs angular frequency.

"50 Hz" is the ordinary frequency $f$: the voltage completes 50 full cycles per second. The angular frequency is a different quantity: …