Alternating Current NCERT Exemplar Solutions - Class 12 Free PDF

Correct option: (b) 5√3/2 A.

Concept used. A sinusoidal alternating current can be written as i(t) = I_m sin(ω t), where I_m is the peak (maximum) value of the current and ω = 2π f is the angular frequency. The peak value is related to the rms (root-mean-square) value by I_m = √2 I_rms. If the current is zero at t=0 and is rising, then at any later time t the instantaneous current is i(t)=I_m sin(ω t).

1. Compute the angular frequency: ω = 2π f = 2π × 50 = 100π rad/s.
2. Compute the peak current from the rms value: I_m = √2 I_rms = √2× 5 = 5√2 A.
3. Find the phase at t = 1/300 s after the current was zero: ω t = 100π × 1300 = π3 rad.
4. Substitute into i(t)=I_m sin(ω t): i = 5√2 sin(π3) = 5√2· √32 = 5√62 = 5√32 A.
5. Numerical check: 5√1.5≈ 5× 1.2247 ≈ 6.12 A, which lies between 0 and the peak I_m≈ 7.07 A, as expected one-sixth of a cycle in.

Option (b): i = 5√3/2 A≈ 6.12 A.

Correct option: (c) -X_g.

Concept used. Maximum power transfer theorem for AC sources. A source of internal impedance Z_g = R_g + jX_g delivers maximum average power to a load impedance Z_L = R_L + jX_L when Z_L is the complex conjugate of Z_g: R_L = R_g, X_L = -X_g. The reason: the reactive parts cancel in the series loop, so the total impedance becomes purely resistive (R_g + R_L = 2R_g), and all the energy that flows around the loop is dissipated only in the resistors.

1. Total impedance of the loop: Z = (R_g + R_L) + j(X_g + X_L).
2. Magnitude squared: |Z|² = (R_g+R_L)² + (X_g+X_L)².
3. Average power delivered to the load resistor R_L: P = I_rms² R_L = V_rms² R_L(R_g+R_L)² + (X_g+X_L)².
4. For a given source V_rms and given R_L, P is largest when the denominator is smallest. The reactance term (X_g+X_L)² ≥ 0, so it is minimised when X_g+X_L = 0, i.e. X_L = -X_g.
5. Why -X_g and not +X_g: X_g might be inductive (+ω L); to cancel it the load must contribute an equal capacitive reactance (-1/ω C). The minus sign is the cancellation.

Option (c): X_L = -X_g.

Correct option: (c).

Concept used. An AC voltmeter cannot show the instantaneous voltage (which oscillates 100 times per second for 50 Hz mains); its needle has too much inertia. Instead, the meter responds to a thermal or rectified effect that averages v² over time. The reading is then converted to the root-mean-square voltage V_rms = √v² , which equals V_m/√2 for a sinusoidal supply.

1. For sinusoidal mains, v(t) = V_m sin(ω t). Its time-average over one cycle is zero, so a true ``average voltmeter'' would read 0.
2. The squared voltage v²(t) = V_m² sin²(ω t) has a non-zero average v² = V_m²/2.
3. Taking the square root: √v² = V_m/√2 ≡ V_rms. The meter is internally calibrated so that the needle position labelled ``220 V'' corresponds to this rms value.
4. Eliminating the other options:
   • (a) is wrong: 220 V is a steady reading because rms is itself steady, not because the input is DC.
   • (b) is wrong: 220 V is the rms value; the peak voltage is V_m = 220√2≈ 311 V.
   • (d) is wrong: a steady reading is expected behaviour for an AC voltmeter, not a fault.

Option (c): the meter reads √v² , i.e. the rms voltage.

Correct option: (b).

Concept used. In a series LCR circuit the resonant angular frequency is ₀ = 1√LC, f₀ = 12π√LC. The resonant frequency is a property of the circuit elements L and C alone; it does not depend on the generator's frequency. To reduce f₀, we must increase either L or C (or both).

1. Examine each option and ask whether it changes L or C, and in which direction:
   • (a) Changing the generator frequency changes the driving frequency, not the resonant frequency of the circuit. Wrong.
   • (b) Adding a capacitor in parallel to the existing one: parallel capacitors add, so C_new = C₁ + C₂ > C₁. This increases C, which decreases ₀ = 1/√LC. Correct.
   • (c) Removing the iron core of the inductor reduces the relative permeability and hence reduces L. That increases ₀. Wrong direction.
   • (d) Removing the dielectric reduces the permittivity and hence reduces C. That also increases ₀. Wrong direction.
2. Conclusion: only (b) increases C, which lowers the resonant frequency.

Option (b): adding a capacitor in parallel increases C and so lowers f₀.

Correct option: (c) R = 15 Ω, L = 3.5 H, C = 30 μF.

Concept used. ``Better tuning'' means the resonance curve must be sharper: only frequencies very close to ₀ should produce a large current. The sharpness is measured by the quality factor Q = 1R√LC. The higher the Q, the narrower the resonance peak. So we should pick the option that maximises Q.

1. Compute Q for each option. Use Q = (1/R)√L/C with C in farads.
   [2pt] (a) √L/C = √1.5/(35× 10^-6) = √4.286× 10⁴ ≈ 207.0. Then Q = 207.0/20 ≈ 10.4.
   [2pt] (b) √L/C = √2.5/(45× 10^-6) = √5.556× 10⁴ ≈ 235.7. Then Q = 235.7/25 ≈ 9.4.
   [2pt] (c) √L/C = √3.5/(30× 10^-6) = √1.167× 10⁵ ≈ 341.6. Then Q = 341.6/15 ≈ 22.8.
   [2pt] (d) √L/C = √1.5/(45× 10^-6) = √3.333× 10⁴ ≈ 182.6. Then Q = 182.6/25 ≈ 7.3.
2. Comparing: Q_c ≈ 22.8 > Q_a ≈ 10.4 > Q_b ≈ 9.4 > Q_d ≈ 7.3. Option (c) wins by a wide margin.
3. Physical intuition: option (c) has the smallest R and the largest L/C ratio, so dissipation is lowest and reactive ``energy reservoir'' is greatest.

Option (c): Q≈ 22.8, the sharpest tuning.

Correct option: (c) 14.4 W.

Concept used. Average power dissipated in an AC circuit equals the power dissipated in the resistive part only (inductors and capacitors do not consume average power). Thus P̄ = I_rms² R, where I_rms = V_rms/Z and Z = √R² + X_L² for an R–L series circuit.

1. Compute the impedance: Z = √R² + X_L² = √2² + 1² = √5 Ω.
2. Compute the rms current: I_rms = V_rmsZ = 6√5 A.
3. Compute the average power: P̄ = I_rms² R = (6√5)²× 2 = 365× 2 = 725 = 14.4 W.
4. Sanity check: if there were no inductor, the power would have been V_rms²/R = 36/2 = 18 W. Adding any reactance must reduce the current and hence the power; 14.4 < 18 is consistent.

Option (c): P̄ = 14.4 W.

Correct option: (a) 1/√2 A.

Concept used. For a resistive load (an incandescent bulb is essentially resistive) at rms voltage V_rms drawing rms current I_rms, the average power is P = V_rms I_rms. The peak current is related to the rms current by I_m = √2 I_rms.

1. Find rms current from the power rating and rms voltage: I_rms = PV_rms = 1224 = 0.5 A = 12 A.
2. Convert to peak current: I_m = √2 I_rms = √2· 12 = √22 = 1√2 A.
3. Numerical: 1/√2≈ 0.707 A.

Option (a): I_m = 1/√2 A≈ 0.707 A.

Correct option: (d).

Concept used. In a series circuit driven by an AC source, the rms current is I_rms = V_rmsZ, Z = √R² + (ω L - 1ω C)². The behaviour with frequency depends on what elements are present:

• Pure R: Z = R is constant, so I does not change with ω.
• Series R–L: Z = √R² + ω² L² grows monotonically with ω, so I only decreases.
• Series R–C: Z = √R² + 1/(ω C)² shrinks monotonically as ω grows, so I only increases.
• Series R–L–C: Z first decreases (capacitor dominates), reaches a minimum Z=R at resonance ₀ = 1/√LC, then increases (inductor dominates). Hence I first increases, then decreases.

1. Match the observed pattern (rise then fall) with the four candidates. Only the series R–L–C combination has this peak-and-drop shape.
2. Option (a), inductor and capacitor (no resistor), would show |Z|=|ω L - 1/(ω C)|. This drops to zero exactly at ₀, where the current becomes infinite, then rises again; a singular sharp spike, not a smooth rise-and-fall.
3. Option (b): R–L gives a monotonically decreasing I with ω. Wrong shape.
4. Option (c): R–C gives a monotonically increasing I with ω. Wrong shape.
5. Only option (d) reproduces the qualitative ``rise to a peak, then fall'' behaviour observed.

Option (d): resistor, inductor and capacitor.

Correct options: (c) and (d).

Concept used. The current rises with frequency only when the impedance falls with frequency. Among the basic elements:

• X_R = R does not depend on ω.
• X_L = ω L increases with ω.
• X_C = 1/(ω C) decreases with ω.

1. (a) Pure R: I = V/R is independent of ω. Wrong.
2. (b) Series R–L: Z = √R² + (ω L)² grows with ω, so I falls. Wrong.
3. (c) Series R–C: Z = √R² + 1/(ω C)² shrinks as ω grows, so I rises. Correct.
4. (d) Pure C: Z = 1/(ω C) shrinks with ω, so I = Vω C grows linearly. Correct.

Options (c) and (d).

Correct options: (a), (b) and (d).

Concept used. For a fixed power P to be delivered, the line current is I = P/V. So raising V lowers I. The power dissipated in the line resistance is P_loss = I² R_line. Therefore a low current means much smaller I² R losses. Voltages are routinely stepped up at the generator and stepped down at the consumer using transformers.

1. (a): With P fixed, I = P/V. Increasing V from say 220 V to 11 kV reduces I by a factor of 50. Correct.
2. (b): The line dissipation is P_loss = I² R_line. A 50× smaller I gives 2500× smaller losses. Correct.
3. (c): Thinner wires have larger R_line, so naively one would lose more power. The real reason transmission wires are not made arbitrarily thin is mechanical strength, voltage breakdown and skin-effect resistance. So (c) is not the main reason high-V transmission is used. Wrong.
4. (d): Transformers transform AC voltages efficiently. At the load end, a step-down transformer brings 11 kV (or 220 kV) back to the 220 V that homes use. This is a feature of AC, not of DC. Correct.

Options (a), (b), (d).

Correct options: (a), (b) and (c).

Concept used. In a steady-state, sinusoidally driven LCR series circuit, the phase angle ϕ between the source voltage and the resulting current is defined by tanϕ = X_L - X_CR = ω L - 1/(ω C)R. Since R>0 and Z≥ R>0, we have cosϕ = R/Z ∈ (0, 1]. Therefore the average power delivered by the source to the circuit, P = V_rms I_rms cosϕ = I_rms² Zcosϕ, is always ≥ 0.

1. Analyse (a): Because cosϕ = R/Z and both R, Z >0, cosϕ is strictly positive. Hence P≥ 0. Correct.
2. Analyse (b): cosϕ = 0 only when R=0, i.e. for a pure LC circuit. In that ideal limit no energy is transferred from source to oscillator on average; the energy just sloshes between L and C. Correct. (More carefully: for finite but very small R the steady-state average power is positive but tiny.)
3. Analyse (c): Since cosϕ≥ 0, P cannot be negative. The source therefore cannot remove energy on average from the oscillator. Correct.
4. Analyse (d): This contradicts (c). If P<0 were possible, the driving force would drain the oscillator; but this never happens in steady state. Wrong.

Options (a), (b), (c).

Correct options: (c) and (d).

Concept used. For an ideal capacitor across a sinusoidal supply v(t) = V_m sinω t:

• The charge on the plates is q(t) = C v(t) = C V_m sinω t, so q is exactly in phase with v.
• The current is i(t) = dq/dt = C V_mω t = I_m sin(ω t + π/2), so i leads v by 90^∘.
• Average power P = V_rmsI_rmscosϕ with ϕ = π/2, so cosϕ = 0 and P = 0.

1. (a): The maximum voltage between plates is V_m = √2· 220 ≈ 311 V, not 220 V. Wrong.
2. (b): i leads v by π/2 in a pure capacitor; they are not in phase. Wrong.
3. (c): From q = Cv, charge and voltage are in phase at every instant. Correct.
4. (d): A pure capacitor stores energy on one quarter-cycle and returns it on the next. The time-average of vi is zero. Correct.

Options (c), (d).

Correct options: (a) and (d).

Concept used. Domestic mains is sinusoidal AC: v(t) = V_msinω t, i(t) = I_msin(ω t - ϕ). The phase ϕ is set by the load. The household load is a mixed load (resistive bulbs, inductive motors, capacitive electronics), so ϕ lies strictly between -π/2 and +π/2. The quoted ``220 V'' is the rms voltage, not the average voltage.

1. (a): The time-average of sinω t over one cycle is zero. So the average current is exactly 0. Correct.
2. (b): The time-average of v(t) is also 0 (not 220 V). The number 220 V is V_rms. Wrong.
3. (c): ϕ = 90^∘ would require a purely reactive load (no resistance). Real household loads always include some resistance, so |ϕ|<π/2. Wrong.
4. (d): For any combination of R, L, C with R>0, cosϕ = R/Z >0 implies |ϕ|<π/2. Correct.

Options (a), (d).

Concept used. In a mass–spring oscillator, the kinetic energy 12 mẋ² is associated with motion (the velocity x), and the potential energy 12 k x² is associated with displacement (the position x). In an LC circuit the analogous quantities are q x, i = q x, 1C k, L m. So inductive energy plays the role of kinetic energy and capacitive energy plays the role of potential energy.

1. Write the two energies for the LC circuit: U_C = q²2C (stored in the capacitor), U_L = 12Li² (stored in the inductor).
2. Compare term by term with 12 kx² and 12 mx²: U_C = 12(1C) q² ↔ 12kx² (PE), U_L = 12L i² ↔ 12mx² (KE).
3. Therefore the capacitor's electric-field energy is analogous to spring potential energy, and the inductor's magnetic-field energy is analogous to the block's kinetic energy.

U_C = q²/2C acts like PE, U_L = Li²/2 acts like KE.

Concept used. For an inductor X_L = ω L and for a capacitor X_C = 1/(ω C). In the high-frequency limit (ω→∞):

• Every inductor behaves like an open circuit (its reactance is huge).
• Every capacitor behaves like a short circuit (its reactance is negligible).

1. Top arm R₁ – C₁ – L₁: the L₁ is an open, so no current flows through the top arm.
2. Bottom arm L₂ – R₂ – R₃: the L₂ is an open, so the left end of the bottom arm is disconnected from the source's left terminal.
3. The remaining conducting path from the source's left terminal to its right terminal is: through R₁, through the shorted C₁ (zero impedance), to the central node, through the shorted C₂ (zero impedance) to the bottom arm, through R₂, then R₃, back to the source.
4. Equivalent circuit therefore reduces to R₁, R₂ and R₃ in series across the source. Inductors are removed (open) and capacitors collapse to wires.

Z_eff = R₁ + R₂ + R₃.

Concept used. In circuit (a) only a resistor is present, so the impedance is Z_a = R and the rms current is I_a = V_rmsR. In circuit (b) we have R, C and L in series, so Z_b = √R² + (ω L - 1ω C)², I_b = V_rmsZ_b. Since Z_b² = R² + (X_L-X_C)² ≥ R², always Z_b≥ R and so I_b ≤ I_a.

1. Part (a), when are the currents equal? I_a = I_b requires Z_b = R, which forces ω L - 1ω C = 0 ⇒ ω = ₀ = 1√LC. That is, the source must drive the circuit at its resonance frequency.
2. Part (b), can I_b > I_a? Since Z_b ≥ R = Z_a for every ω, the current in (b) is at most equal to that in (a). It can never exceed it.

(a) At resonance, ω = 1/√LC.   (b) No, I_b can never exceed I_a.

Concept used. The instantaneous power delivered by an AC source is p(t) = v(t) i(t). With v = V_msinω t and i = I_msin(ω t - ϕ), p(t) = V_m I_m sinω t sin(ω t - ϕ) = 12 V_m I_m[cosϕ - cos(2ω t - ϕ)]. The average power over a full cycle is P = 1T₀^T p(t) dt = 12 V_m I_mcosϕ = V_rms I_rmscosϕ.

1. Instantaneous: p(t) has an oscillating term cos(2ω t - ϕ) whose amplitude is 12 V_m I_m. During parts of the cycle this term exceeds the constant 12 V_m I_mcosϕ, so p(t) < 0 during those intervals. Physically, this is the time when the inductor/capacitor returns energy to the source. So yes, instantaneous power can be negative.
2. Average: For a passive load the phase ϕ satisfies cosϕ = R/Z ≥ 0, so P ≥ 0. The source delivers (or at worst, breaks even with) the load over a full cycle. So no, average power cannot be negative for any passive AC circuit.

Instantaneous power: yes, can be negative; average power: no, cannot be negative.

Concept used. The bandwidth ω of a resonance curve is defined as the difference between the two frequencies ₁ < ₀ < ₂ at which the current drops to I_max/2 (the half-power points, since power ∝ I² drops to 12 at those frequencies). Then ω = ₂ - ₁.

1. Read the peak current from Fig. 7.3: I_max = 1.0 A at ₀ = 1.0 rad/s.
2. Half-power level: I_max/2 = 1.0/2 ≈ 0.707 A.
3. Drop horizontally from this I value onto the curve and read off the two intersection points: ₁ ≈ 0.8 rad/s on the left and ₂ ≈ 1.2 rad/s on the right.
4. Bandwidth: ω = ₂ - ₁ = 1.2 - 0.8 = 0.4 rad/s.

ω ≈ 0.4 rad/s (the band between ₁≈ 0.8 and ₂≈ 1.2 rad/s).

Concept used. The rms current is defined by I_rms = √i² = √1T₀^T i²(t) dt. For a piecewise-constant waveform that takes value i_k for a fraction f_k = t_k/T of the period (with ∑ f_k = 1), this simplifies to I_rms = √_k f_k i_k².

1. Read Fig. 7.4: the current is +1 A for half the period and -2 A for the other half. So i₁=+1 A with f₁=1/2, and i₂=-2 A with f₂=1/2.
2. Mean of i²: i²= 12 (1)² + 12 (-2)² = 12 + 12· 4 = 12+2 = 52.
3. Take the square root: I_rms = √52 = √2.5 ≈ 1.58 A.
4. Marking the result: draw a horizontal line at I = +1.58 A across the entire t-axis on Fig. 7.4 (and a symmetric one at -1.58 A would also represent the same rms value, since rms is sign-blind).

I_rms = √5/2 A≈ 1.58 A, marked as a horizontal dashed line on Fig. 7.4.

Concept used. In a series LCR circuit driven by an AC source, the source voltage leads the current by an angle ϕ given by tanϕ = X_L - X_CR = ω L - 1/(ω C)R. The sign of ϕ depends on the relative size of X_L and X_C, which themselves depend on ω.

1. Very low ω: X_L = ω L is small, X_C = 1/(ω C) is large. So X_L - X_C < 0, giving tanϕ < 0, i.e. ϕ is negative (the voltage actually lags the current; equivalently the current leads the voltage).
2. At resonance, ω = ₀ = 1/√LC: X_L = X_C, so tanϕ = 0 and ϕ = 0. Voltage and current are in phase.
3. Very high ω: X_L is large, X_C is small. So X_L - X_C > 0, tanϕ > 0, and ϕ is positive (voltage genuinely leads current).
4. As ω increases from very low to very high, ϕ moves continuously from a negative value, through 0 at ₀, to a positive value.

ϕ starts negative at low ω, passes through ϕ = 0 at the resonance frequency ₀ = 1/√LC, then becomes positive at high ω.

Concept used. For an AC source v(t) = V_m sinω t feeding a device that draws a current i(t) = I_m sin(ω t ± π/2) (i.e. exactly 90^∘ out of phase with the voltage), the instantaneous power is p(t) = v(t) i(t) = V_m I_m sinω tcosω t = 12 V_m I_m sin 2ω t. Notice three things: (i) p(t) oscillates at twice the source frequency; (ii) it is positive on alternate quarter cycles and negative on the others; (iii) its average over a full cycle is zero. Among the three sinusoids in Fig. 7.5, the one that oscillates at twice the others' frequency and whose mean is zero is the power curve.

1. Examine Fig. 7.5. Curves A and B have the same period T, but B's peak is shifted from A's by a quarter cycle (a π/2 phase difference). Curve C completes two oscillations in the time A and B complete one, and is centred at zero.
2. Part (a): The power p(t) ∝ sin 2ω t oscillates at twice the source frequency. So curve C is the power curve.
3. Part (b): Compute the average: P = 1T₀^T 12 V_m I_m sin 2ω t dt = 0. Equivalently, P = V_rmsI_rmscosϕ with ϕ = π/2, so cosϕ = 0 and P = 0.
4. Part (c): A device with ϕ = ±π/2 between v and i is a purely reactive element: either an ideal inductor (current lags voltage by π/2) or an ideal capacitor (current leads voltage by π/2). From the standard convention shown in Fig. 7.5 (current peaks before voltage), i leads v, so X is an ideal capacitor.

(a) Curve C.   (b) P = 0 over a cycle.   (c) Device X is a pure capacitor (or equally well a pure inductor; in either case, a wattless element).

Concept used. For a steady direct current, the ampere is the rate of flow of charge: 1 A = 1 C/s. For an alternating current the value oscillates and the simple ``rate of charge flow'' would average to zero. The standard convention is therefore to define the AC ampere by its heating equivalence to a DC current.

1. Pass an alternating current i(t) through a fixed resistor R. The instantaneous power dissipated is p(t) = i²(t) R, and the heat liberated per cycle is ₀^T i²(t) R dt.
2. Now pass a steady DC current I_eq through the same resistor. In one period it liberates I_eq² R T of heat.
3. Demand that the two heats be equal: I_eq² R T = R₀^T i²(t) dt ⇒ I_eq = √1T₀^T i²(t) dt.
4. The right-hand side is precisely the rms current I_rms. So the AC ampere is the rms value of the alternating current, defined as the steady direct current that would produce the same heating in the same resistor over one cycle.
5. For a sinusoid i(t) = I_msinω t, I_rms = I_m/2.

The ampere for an AC is the rms value I_rms: the steady DC that delivers the same average heat in a resistor over one cycle as the alternating current does.

Concept used. A real coil is modelled as an inductor L in series with its own internal resistance R. Across an AC supply of angular frequency ω = 2π f, the impedance is Z = √R² + X_L², X_L = ω L, and the current lags the voltage by a phase angle ϕ such that tanϕ = X_LR. The corresponding time lag is Δ t = ϕ/ω.

1. Compute ω: ω = 2π f = 2π × 50 = 100π rad/s.
2. Compute the inductive reactance: X_L = ω L = 100π × 0.01 = π Ω ≈ 3.14 Ω.
3. Compute the impedance: Z = √R² + X_L² = √1² + π² = √1 + 9.8696 = √10.8696 ≈ 3.30 Ω.
4. Compute the phase angle: tanϕ = X_LR = π1 = π ≈ 3.14, ϕ = arctanπ ≈ 72.34^∘ = 72.34 × π180 rad ≈ 1.263 rad.
5. Compute the time lag: Δ t = ϕω = 1.263100π = 1.263314.16≈ 4.02× 10^-3 s.

Z ≈ 3.30 Ω; time lag Δ t ≈ 4.0× 10^-3 s = 4 ms.

Concept used. An ideal transformer obeys V_sV_p = N_sN_p = I_pI_s, and conserves average power: P_p = P_s, so V_p I_p = V_s I_s. ``Line voltage'' in India is the standard V_p = 220 V (rms).

1. Find the secondary voltage from the load. For a resistive (or nearly resistive) load P_s = V_s I_s, so V_s = P_sI_s = 600.54 V ≈ 111.1 V.
2. Apply power conservation (no losses, so P_p = P_s = 60 W): I_p = P_pV_p = 60220 A ≈ 0.27 A.
3. Check via the turns ratio: V_s/V_p = 111.1/220 ≈ 0.505, and I_p/I_s = 0.27/0.54 = 0.5. The two ratios agree (within rounding), so the relation V_s/V_p = I_p/I_s holds, confirming an ideal transformer.
4. Since V_s < V_p (111 V < 220 V), the transformer steps the voltage down.

Primary current I_p ≈ 0.27 A; the transformer is a step-down type.

Concept used. For a sinusoidal voltage v(t) = V_msinω t across a capacitor C, the charge stored on its plates is q(t) = Cv(t), and the current flowing into/out of the plates is i(t) = dqdt = C dvdt = CV_mω t. So the peak current is I_m = CV_mω, and the ratio V_m/I_m ≡ X_C is the capacitive reactance: X_C = V_mI_m = 1ω C. This is inversely proportional to ω.

1. Mathematically: from X_C = 1/(ω C), as ω ↑, X_C ↓. At very high frequencies X_C → 0 (capacitor looks like a short circuit); at ω → 0 (DC), X_C → ∞ (open circuit).
2. Physically: a capacitor charges and discharges every half-cycle. At low frequencies the voltage spends a long time near its peak, giving the plates plenty of time to fully charge; the back-voltage q/C across the plates rises to match the source, choking off the current. At high frequencies the voltage reverses before much charge accumulates, so the back-voltage never builds up. The source therefore sees a small effective impedance.
3. Equivalently, the impedance is the ratio of peak voltage to peak current: X_C = V_m/I_m. Since I_m = Cω V_m grows linearly with ω, the ratio falls as 1/ω.

X_C = 1/(ω C): higher ω leaves less time for the plates to charge up, so the current is bigger and the effective opposition is smaller.

Concept used. An inductor L produces a back-EMF whenever the current through it changes: ε = -L di/dt (Faraday–Lenz law). For a sinusoidal current i(t) = I_msinω t, the back-EMF is ε = -Ldidt = -L I_mω t. Its peak amplitude is V_m = L I_m ω, and the ratio V_m/I_m ≡ X_L is the inductive reactance: X_L = ω L.

1. Mathematically: from X_L = ω L, as ω↑, X_L↑. At ω → 0 (DC) X_L→ 0 (inductor looks like a wire); at high frequencies X_L→∞ (inductor looks like an open break).
2. Physically: the current must oscillate at the source frequency, so di/dt is large at high ω. A larger di/dt means a larger induced back-EMF L di/dt, which opposes the change in current more strongly. Hence the inductor presents more opposition to current at higher ω.
3. Equivalently, X_L = V_m/I_m = L I_mω/I_m = ω L, growing linearly with ω.

X_L = ω L: a faster-changing current produces a bigger back-EMF, so the inductor opposes the current more.

Concept used. For an AC circuit driven at V_rms, drawing I_rms with the current lagging the voltage by phase ϕ, the average power, impedance and components are P = V_rms I_rms cosϕ, Z = V_rmsI_rms, R = Zcosϕ, X = Z|sinϕ|. The peak current is I_M = 2 I_rms. Treating the magnitudes alone, |tanϕ| = 3/4 gives sinϕ = 3/5 and cosϕ = 4/5 (the familiar 3–4–5 triangle).

1. Given: P = 2000 W, V_rms² = 50,000 V² (so V_rms≈ 223.6 V), cosϕ = 4/5 = 0.8, sinϕ = 3/5 = 0.6.
2. Find the impedance from the power relation P = V_rms²/Z (the convention used by the NCERT Exemplar): Z = V_rms²P = 50,0002000 = 25 Ω.
3. Find I_rms: I_rms = V_rmsZ = 223.625 ≈ 8.944 A.
4. (i) Resistance: R = Zcosϕ = 25 × 45 = 20 Ω.
5. (ii) Net reactance: since the current lags, the net reactive part is inductive, so X_L > X_C and |X_C - X_L| = Z|sinϕ| = 25×35 = 15 Ω. Therefore X_C - X_L = -15 Ω (or equivalently X_L - X_C = +15 Ω). Cross-check: R² + (X_L-X_C)² = 400 + 225 = 625 = 25² = Z².
6. (iii) Peak current: I_M = 2 I_rms = 2 × 8.944 ≈ 12.65 A≈ 12.6 A.
7. Second device: if R → 2R, X_L → 2X_L, X_C → 2X_C, every impedance doubles, but tanϕ = (X_L-X_C)/R is unchanged, so ϕ is the same. Then Z → 2Z = 50 Ω, I_rms → V/(2Z) = I_rms/2 ≈ 4.47 A, and I_M → I_M/2 ≈ 6.32 A. Power P → V²/(2Z) = 1000 W (halved).

(i) R = 20 Ω; (ii) |X_C - X_L| = 15 Ω; (iii) I_M ≈ 12.6 A.   If R, X_C, X_L all double: ϕ unchanged, Z→ 50 Ω, I_M→ 6.32 A, P→ 1 kW.

Concept used. For a load drawing power P at line voltage V, the line current is I = P/V. The total resistance of the transmission line (two wires, length each, cross-section A) is R_line = 2ρA. The ohmic loss in the line is P_loss = I² R_line, so the fraction of power lost is P_lossP = I² R_lineP = P R_lineV². Larger V ⇒ smaller fraction (quadratically). This is why electricity is transmitted at high voltage.

1. Compute line resistance once and for all. Cross-section: A = π r² = π (0.5× 10^-2)² = π× 2.5× 10^-5 m² ≈ 7.854× 10^-5 m². Total length of conductor (two wires of 10 km each): 2= 2 × 10 000 = 20 000 m. Resistance: R_line = ρ (2)A = (1.7× 10^-8)(2× 10⁴)7.854× 10^-5 = 3.4× 10^-47.854× 10^-5 ≈ 4.33 Ω.
2. (i) Transmission at 220 V. Line current: I = PV = 10⁶220 ≈ 4545 A. Power loss: P_loss = I² R_line = (4545)² × 4.33 = 2.066× 10⁷ × 4.33 ≈ 8.95× 10⁷ W. Fraction: P_lossP = 8.95× 10⁷10⁶ ≈ 89.5. That is, the loss is 89.5 times the power being delivered to the town. It is wildly infeasible: 89.5 MW of heat would be dumped in the wires just to deliver 1 MW. The wires would melt instantly.
3. (ii) Transmission at 11,000 V. Line current: I = PV = 10⁶11,000 ≈ 90.9 A. Power loss: P_loss = I² R_line = (90.9)² × 4.33 = 8264 × 4.33 ≈ 3.58× 10⁴ W ≈ 35.8 kW. Fraction: P_lossP = 3.58× 10⁴10⁶ ≈ 0.0358 ≈ 3.58%. Only about 3.6% of the transmitted power is lost as heat. This is entirely feasible.
4. Comparison. Raising the voltage by a factor of 50 (from 220 to 11,000 V) shrinks the loss fraction by 50² = 2500. This is the entire reason power grids step the voltage up at the generating station and step it back down at the consumer.

(i) P_loss/P ≈ 89.5 at 220 V, completely infeasible.   (ii) P_loss/P ≈ 3.6% at 11,000 V, feasible.

Concept used. The circuit shows a resistor R in series with a parallel combination of an inductor L and a capacitor C, driven by v = v_msinω t. For a parallel LC block, the voltage across both branches is the same, while the currents add. For an inductor the current lags the voltage by π/2; for a capacitor the current leads the voltage by π/2. The reactances are X_L = ω L and X_C = 1/(ω C).

1. Let the voltage across the parallel LC block be v_LC = v_m'sinω t (we will determine the amplitude v_m' shortly). The two branch currents are i_L = v_m'ω Lsin(ω t - π2), i_C = v_m' ω C sin(ω t + π2).
2. Since the two reactive currents are π out of phase with each other, their sum is i_LC = i_L + i_C = v_m'(ω C - 1ω L)sin(ω t + π2). Let B ≡ ω C - 1/(ω L). Then i_LC = v_m' Bsin(ω t + π/2).
3. The same current i_LC also flows through the series resistor R, so the net current in the source loop is i = i_LC.
4. KVL around the loop: v = iR + v_LC. Writing in phasors (using v_LC = i/(jB) for the parallel block): v = R i + 1jB i = i(R - jB). Therefore i = vR - j/B = vZ, with Z = R - jB = R - jω C - 1/(ω L), |Z| = √R² + 1(ω C - 1/ω L)².
5. Magnitude of the current: I_m = v_m|Z| = v_m√R² + 1(ω C - 1/ω L)².
6. Phase of i relative to v: from i = v/Z, tanϕ = Im(Z)Re(Z) = -1/BR = -1R (ω C - 1/ω L). If ω C > 1/(ω L) (the parallel block is capacitive), B>0 and ϕ<0, so i leads v. Otherwise i lags.
7. We have shown i = v/Z, i.e. Ohm's law for the AC circuit, with Z as quoted.

Z = √R² + 1(ω C - 1/(ω L))²; phase tanϕ = -1/[R(ω C - 1/ω L)]; i = v/Z.

Concept used. The terms 12 Li² and q²/(2C) are the magnetic and electric energies stored in the inductor and capacitor; Ri² is the resistive dissipation rate; vi is the instantaneous power supplied by the source. We will use i = dq/dt to convert i(di/dt) into a total time-derivative of i²/2, and iq = q dq/dt into the time-derivative of q²/2.

1. (i) Multiply through by i: Lididt + Ri² + q iC = v_m isinω t. Using i = dq/dt, i di/dt = 12 d(i²)/dt and qi/C = (1/C) q dq/dt = 12Cd(q²)/dt. So the equation becomes ddt[12 L i²] + Ri² + ddt[q²2C] = v_m i sinω t. Group the energy-derivatives: ddt[12Li² + q²2C] + Ri² = v_m i sinω t.
2. (ii) Physical interpretation of each term:
   • 12 L i²: magnetic energy stored in the inductor. Its time derivative is the rate at which the inductor accumulates magnetic energy.
   • q²/(2C): electric energy stored in the capacitor. Its derivative is the rate of electric-energy accumulation in C.
   • Ri²: power dissipated by the resistor (Joule heating).
   • v_m i sinω t = v(t) i(t): instantaneous power supplied by the source.
3. (iii) Conservation of energy: ddt[12 Li² + q²2C]_{rate of change of stored energy} + Ri²_{rate of dissipation} = v i_{rate of source supply}. In words: the power supplied by the source equals the rate of energy storage in L and C plus the rate of heat dissipation in R. This is an energy-balance/continuity equation for the circuit.
4. (iv) Integrate over one complete cycle of period T: In steady state, i and q are periodic with period T, so 12 L i² and q²/(2C) each return to their starting values. Hence ₀^T ddt[12 Li² + q²2C] dt = 0. The cycle-integrated energy-balance becomes ₀^T Ri² dt = ₀^T v(t) i(t) dt. The left side is Ri²T = R I_rms² T > 0 (provided R>0 and the current is non-zero), so the right side must also be positive: ₀^T v i dt = V_m I_m T · 12cosϕ = V_rms I_rms T cosϕ > 0. Therefore cosϕ > 0, which means |ϕ| < π/2. The phase difference is acute.

(i)–(iii) The product v i = d(12 Li² + q²/(2C))/dt + Ri² is the circuit's energy conservation. (iv) Integrating over a cycle gives cosϕ > 0, so |ϕ| < π/2.

Concept used. In a series LCR circuit driven by a source v = v_msinω t, Kirchhoff's voltage law applied around the loop gives L didt + Ri + qC = v. Using i = dq/dt, this becomes the equation of motion for the charge q(t). Once the source is removed and the resistor is shorted, the remaining circuit is a pure LC oscillator with natural angular frequency ₀ = 1/√LC.

1. (i) Equation of motion for q(t). Substitute i = dq/dt and di/dt = d² q/dt² into Kirchhoff's law: L d² qdt² + R dqdt + qC = v_msinω t. This is a forced, damped harmonic-oscillator equation for q, with ``mass'' L, ``damping'' R, ``spring constant'' 1/C, and a sinusoidal driving force v_msinω t.
2. (ii) Energies stored at t = t₀. Let i₀ be the current and q₀ the charge at that instant. Then U_L(t₀) = 12L i₀², U_C(t₀) = q₀²2C. These are the magnetic energy stored in L and the electric energy stored in C respectively.
3. (iii) Subsequent motion. After t = t₀: the source v is removed, and R is short-circuited (a wire replaces R). The circuit reduces to a closed loop containing only L and C. Kirchhoff's voltage law now gives L d² qdt² + qC = 0 ⇒ d² qdt² = -₀² q, with ₀ = 1/√LC. The general solution is q(t) = q₀ cos[₀ (t-t₀)] + i₀₀sin[₀ (t-t₀)], a pure sinusoidal oscillation at the natural frequency. Physically: the capacitor discharges through the inductor, building up current; once q→ 0 the inductor's stored magnetic energy reverses the current, recharging the capacitor with the opposite polarity. With no resistance, no energy is lost, so the oscillation continues with constant total energy U_tot = 12 L i₀² + q₀²2C = const.

(i) Lq + Rq + q/C = v_msinω t. (ii) U_L = 12 L i₀², U_C = q₀²/(2C). (iii) Undamped LC oscillation at ₀ = 1/√LC; total energy 12 L i₀² + q₀²/(2C) conserved.