Which quantity cannot change instantaneously in a circuit containing finite voltages?

the current through an inductor

the current through a capacitor

Which statements about the reciprocity theorem are correct?

it applies to linear, bilateral networks excited by a single independent source

interchanging an ideal voltage source and an ideal ammeter leaves the ammeter reading unchanged

it applies to linear, bilateral networks excited by a single independent source

it holds for every network containing dependent sources

interchanging an ideal voltage source and an ideal ammeter leaves the ammeter reading unchanged

it applies to circuits containing diodes

When applying superposition, an ideal independent current source is deactivated by replacing it with

its internal resistance in series

For a parallel RLC circuit with ideal elements driven by a current source, which statements hold at resonance?

the impedance seen by the source is at its maximum

the resonant frequency is 1/(2π√(LC))

the impedance seen by the source is at its maximum

the source current drawn from a fixed-voltage source would be at its maximum

the resonant frequency is 1/(2π√(LC))

Electrical Circuits: Sources, Network Theorems, Transients, AC Power, Resonance and Bode Plots

Section 2 of the GATE Biomedical Engineering (BM) paper, Electrical Circuits, in one chapter and in the syllabus’s own order. It starts with sources — independent and dependent, ideal and practical — and the v-i relations of the resistor, inductor and capacitor. Then Kirchhoff’s laws and the five named theorems: superposition, Thevenin, Norton, maximum power transfer and reciprocity. Then the transient response of RC, RL and RLC circuits to dc excitation. Then ac quantities — peak, average and rms values, apparent, active and reactive power — and phasor analysis with impedance and admittance. It closes with series and parallel resonance, the basic filters that R, L and C elements realise, and the Bode plot that describes them. Every one of these ideas returns later in the paper: the RC low-pass is the first stage of an ECG amplifier, and the rms value is what a mains-leakage limit is written in.

1. Sources and the three passive elements

An ideal voltage source holds its terminal voltage whatever current it delivers; an ideal current source holds its current whatever voltage appears across it. A practical voltage source is an ideal one in series with an internal resistance R_s, so its terminal voltage falls as V = V_s − IR_s; a practical current source is an ideal one in parallel with R_p. The two practical models are interchangeable (source transformation) with V_s = I_s R and the same R. A dependent source is set by a voltage or current elsewhere in the circuit — voltage-controlled voltage source (VCVS), voltage-controlled current source (VCCS), current-controlled voltage source (CCVS), current-controlled current source (CCCS) — and it is how transistors and op-amps enter circuit analysis.

The three passive elements
Element	v-i relation	Stored energy	What cannot jump
Resistor R	v = Ri	none — dissipates p = i²R	nothing
Inductor L	v = L di/dt	½Li²	current i_L
Capacitor C	i = C dv/dt	½Cv²	voltage v_C

At dc steady state di/dt = 0 and dv/dt = 0, so an inductor behaves as a short circuit and a capacitor as an open circuit. At the instant of switching the opposite reading applies to an uncharged element: a capacitor with v_C = 0 looks like a short and an inductor with i_L = 0 looks like an open circuit, because neither can change its state variable in zero time without infinite power.

2. Kirchhoff’s laws and the network theorems

KCL: the currents leaving any node sum to zero (charge conservation). KVL: the voltages around any closed loop sum to zero (energy conservation). Node analysis writes KCL at each non-reference node in terms of node voltages; mesh analysis writes KVL around each mesh in terms of mesh currents. The theorems below all rest on linearity.

The five theorems the section names
Theorem	Statement	Where it needs care
Superposition	Response = sum of responses to each independent source acting alone	kill a voltage source by a short, a current source by an open; dependent sources stay; power does not superpose
Thevenin	Any linear two-terminal network = V_th in series with R_th	V_th is the open-circuit voltage; with dependent sources find R_th by a test source, or as V_oc/I_sc
Norton	Same network = I_N in parallel with R_N	I_N is the short-circuit current; R_N = R_th, I_N = V_th/R_th
Maximum power transfer	R_L = R_th gives P_max = V_th²/(4R_th)	for ac, Z_L = Z_th* (the complex conjugate); efficiency at this point is only 50%
Reciprocity	In a linear bilateral network with one source, swapping the source and the ammeter (or voltmeter) positions leaves the reading unchanged	fails with dependent sources in general, and with more than one independent source

⚠️ Deactivating a source is not deleting it

Setting an ideal voltage source to zero leaves a wire, not a gap; setting a current source to zero leaves a gap, not a wire. Swapping these two is the commonest superposition and Thevenin error, and it changes R_th completely.

3. Transients with dc excitation

A first-order circuit (one energy-storing element after reduction) has every voltage and current of the form x(t) = x(∞) + [x(0⁺) − x(∞)] e^−t/τ, with τ = RC for an RC circuit and τ = L/R for an RL circuit, R being the Thevenin resistance seen by the element. Three numbers settle the whole waveform: the initial value from continuity (v_C or i_L), the final value from the dc steady state, and τ. After one time constant the response has covered 63.2% of its change, after 5τ more than 99%.

A series RLC circuit obeys a second-order equation with characteristic roots s = −α ± √(α² − ω₀²), where α = R/(2L) and ω₀ = 1/√(LC); the damping ratio is ζ = α/ω₀ = (R/2)√(C/L). The response is overdamped (two real roots) when α > ω₀, critically damped (a repeated root, the fastest non-oscillating response) when α = ω₀, and underdamped (decaying oscillation at ω_d = √(ω₀² − α²)) when α < ω₀. For a parallel RLC circuit α = 1/(2RC) instead, so a larger parallel resistance means less damping.

🧠 Find τ with the sources killed

The time constant belongs to the circuit, not to the source: deactivate every independent source, find the resistance seen from the capacitor or inductor terminals, and τ = R_eq C or L/R_eq. The same τ then governs every voltage and current in the circuit.

4. AC quantities, phasors and power

The rms value of a waveform is the dc value that would dissipate the same power in a resistor, X_rms = √(mean of x²); the average value of a symmetrical wave is taken over a half cycle. The form factor is rms/average and the crest factor peak/rms.

Standard waveforms, peak value V_m
Waveform	Average	rms	Form factor
Sine (half-cycle average)	2V_m/π = 0.637V_m	V_m/√2 = 0.707V_m	1.11
Half-wave rectified sine	V_m/π	V_m/2	1.57
Square wave ±V_m	V_m	V_m	1
Triangular wave	V_m/2	V_m/√3	1.155

In sinusoidal steady state each quantity becomes a phasor and each element an impedance: Z_R = R, Z_L = jωL, Z_C = 1/(jωC) = −j/(ωC). Impedance Z = R + jX combines like resistance; its reciprocal is the admittance Y = G + jB, which adds in parallel. With V and I as rms phasors and φ the angle by which current lags voltage, active power P = VI cos φ (watts), reactive power Q = VI sin φ (VAR, positive for an inductive load) and apparent power |S| = VI (VA), with the complex power S = VI* = P + jQ and power factor cos φ = P/|S|.

⚠️ P = I²R uses only the resistance

In a series R-X load the active power is I²R and the reactive power is I²X; writing V²/R with the supply voltage across the whole impedance overstates P, because the supply voltage is not across R alone.

5. Resonance, RLC filters and the Bode plot

A series RLC circuit resonates where X_L = X_C, at ω₀ = 1/√(LC) (f₀ = 1/(2π√(LC))). There the impedance is at its minimum, purely resistive and equal to R, the current is at its maximum and in phase with the voltage, and the voltage across L or C is Q times the supply. The quality factor is Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C) and the half-power bandwidth BW = f₀/Q = R/(2πL) in hertz. A parallel RLC tank resonates at the same ω₀ (ideal elements), but there the impedance is at its maximum, equal to R, the source current is at its minimum, and Q = R/(ω₀L) = ω₀CR = R√(C/L).

Basic filters realised with R, L and C
Filter	Realisation	Transfer function	Key frequency
Low-pass	series R, output across C	1/(1 + jωRC)	f_c = 1/(2πRC)
High-pass	series C, output across R	jωRC/(1 + jωRC)	f_c = 1/(2πRC)
Band-pass	series LC, output across R	peak of 1 at ω₀	f₀, BW = f₀/Q
Band-stop (notch)	series R, output across the series LC	zero at ω₀	f₀ — e.g. to reject 50 Hz mains

A Bode plot draws 20 log₁₀|H(jω)| in dB and the phase against log ω. Each simple pole at ω_p adds a slope of −20 dB/decade beyond ω_p and a phase falling from 0 to −90°, passing −45° at ω_p, where the true magnitude is 3 dB below the asymptote; each simple zero does the opposite (+20 dB/decade, +90°). A pole at the origin gives −20 dB/decade everywhere; a complex pole pair gives −40 dB/decade beyond ω_n, with a resonant peak near ω_n when ζ is small. The asymptotic plot is built by adding these straight-line pieces, starting from the low-frequency gain.

🎯 Why Q is both selectivity and voltage magnification

At resonance the inductor and capacitor trade energy back and forth while R removes a little each cycle. Q is 2π times the energy stored over the energy lost per cycle, so a high-Q circuit both rings for many cycles (narrow bandwidth) and builds large voltages across L and C from a small drive.

Key takeaways

Capacitor voltage and inductor current are continuous; at dc steady state C is open and L is short.
Kill voltage sources with shorts and current sources with opens; keep dependent sources; R_th = V_oc/I_sc when dependent sources are present.
First order: x(∞) + [x(0⁺) − x(∞)]e^−t/τ; series RLC: α = R/2L against ω₀ = 1/√LC decides the damping.
P = VI cos φ, Q = VI sin φ, |S| = VI; sine rms = V_m/√2, half-wave rectified rms = V_m/2.
Series resonance: minimum Z, Q = ω₀L/R, BW = f₀/Q; each pole is −20 dB/decade and −45° at its corner.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which quantity cannot change instantaneously in a circuit containing finite voltages?
1. the current through an inductor
2. the voltage across an inductor
3. the current through a capacitor
4. the current through a resistor
Show answer
Answer: A — the current through an inductor
v = L di/dt, so a step change in inductor current would need an infinite voltage; the inductor current is therefore continuous. The inductor voltage, the capacitor current and the resistor current can all jump at a switching instant — it is the capacitor voltage, not its current, that is continuous.
A 12 V ideal source in series with 4 Ω feeds a 12 Ω resistor connected across the output terminals, and a variable load R_L is connected across those same terminals. The maximum power (in W) that can be delivered to R_L is ____.
Numerical answer — type the value.
Show answer
Answer: 6.75
V_th = 12 × 12/(4 + 12) = 9 V and R_th = 4 ∥ 12 = 3 Ω, so with R_L = 3 Ω, P_max = V_th²/(4R_th) = 81/12 = 6.75 W. Using the source voltage 12 V instead of V_th gives 12 W; using the 4 Ω alone as R_th gives 5.06 W.
An uncharged 100 µF capacitor is charged from a 10 V dc source through 10 kΩ, starting at t = 0. The capacitor voltage (in V) at t = 2 s, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 8.65
τ = RC = 10⁴ × 10⁻⁴ = 1 s, so v(2) = 10(1 − e⁻²) = 10 × 0.8647 = 8.65 V. Writing 10e⁻² = 1.35 V gives the voltage across the resistor, not the capacitor; taking τ = 10 s through a factor-of-ten unit slip gives 1.81 V.
A series RLC circuit has R = 2 Ω, L = 1 H and C = 1 F. Its natural response is
1. critically damped
2. overdamped
3. underdamped
4. undamped
Show answer
Answer: A — critically damped
α = R/(2L) = 1 s⁻¹ and ω₀ = 1/√(LC) = 1 rad/s, so α = ω₀ and ζ = 1: a repeated root at s = −1, the critically damped case. The parallel-RLC formula α = 1/(2RC) = 0.25 would wrongly suggest an underdamped response.
A half-wave rectified sine wave has a peak value of 20 V. Its average value (in V), to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 6.37
Averaged over the full period the half-wave rectified sine gives V_m/π = 20/π = 6.37 V. 2V_m/π = 12.73 V is the full-wave value, and V_m/2 = 10 V is the rms of the half-wave wave, not its average.
A 230 V (rms) supply feeds a series load Z = 30 + j40 Ω. The reactive power drawn (in VAR), to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 846.4
|Z| = 50 Ω, so I = 230/50 = 4.6 A and Q = I²X = 21.16 × 40 = 846.4 VAR (the active power is I²R = 634.8 W and the power factor 0.6 lagging). V²/X = 1322.5 VAR wrongly puts the whole supply voltage across the reactance.
A series RLC circuit has L = 10 mH, C = 1 µF and R = 10 Ω. Its half-power bandwidth (in Hz), to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 159.2
ω₀ = 1/√(10⁻² × 10⁻⁶) = 10⁴ rad/s, f₀ = 1591.5 Hz and Q = ω₀L/R = 100/10 = 10, so BW = f₀/Q = 159.2 Hz — equivalently R/(2πL) = 10/(0.0628) = 159.2 Hz. R/L = 1000 is the bandwidth in rad/s; quoting it as hertz is the usual slip.
For H(s) = 100/[(1 + s/10)(1 + s/1000)], the asymptotic Bode magnitude at ω = 100 rad/s is
1. 20 dB
2. 40 dB
3. 0 dB
4. 37 dB
Show answer
Answer: A — 20 dB
The low-frequency gain is 20 log 100 = 40 dB. The pole at 10 rad/s bends the asymptote to −20 dB/decade, so one decade later, at 100 rad/s, it reads 40 − 20 = 20 dB; the second pole at 1000 rad/s has not yet acted. 37 dB is the true value at the first corner, not at 100 rad/s.
Which statements about the reciprocity theorem are correct?
1. it applies to linear, bilateral networks excited by a single independent source
2. it holds for every network containing dependent sources
3. interchanging an ideal voltage source and an ideal ammeter leaves the ammeter reading unchanged
4. it applies to circuits containing diodes
Show answer
Answer: A — it applies to linear, bilateral networks excited by a single independent source; C — interchanging an ideal voltage source and an ideal ammeter leaves the ammeter reading unchanged
Reciprocity needs linearity, bilateral elements and a single source; then swapping the source and the ideal ammeter gives the same reading. Dependent sources generally destroy the symmetry of the network equations, and a diode is neither linear nor bilateral.
When applying superposition, an ideal independent current source is deactivated by replacing it with
1. an open circuit
2. a short circuit
3. its internal resistance in series
4. a dependent source
Show answer
Answer: A — an open circuit
A current source set to zero passes no current whatever the voltage, which is an open circuit. It is the voltage source that becomes a short when set to zero; confusing the two changes the resistances the other sources see.
An RC low-pass filter uses R = 1 kΩ and C = 0.1 µF. Its −3 dB cut-off frequency is approximately
1. 1.59 kHz
2. 10 kHz
3. 159 Hz
4. 628 Hz
Show answer
Answer: A — 1.59 kHz
RC = 10³ × 10⁻⁷ = 10⁻⁴ s, so f_c = 1/(2πRC) = 1591.5 Hz ≈ 1.59 kHz. 10 kHz is 1/RC read as hertz; it is the corner in rad/s.
A 10 V dc source is switched at t = 0 onto a series circuit of R = 100 Ω and L = 0.5 H with no initial current. The current (in mA) at t = 5 ms, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 63.2
τ = L/R = 0.5/100 = 5 ms, so at t = τ, i = (V/R)(1 − e⁻¹) = 100 mA × 0.632 = 63.2 mA. Using τ = RL = 50 s gives almost zero; writing (V/R)e⁻¹ = 36.8 mA gives the decaying, not the rising, current.
For a parallel RLC circuit with ideal elements driven by a current source, which statements hold at resonance?
1. the impedance seen by the source is at its maximum
2. the source current drawn from a fixed-voltage source would be at its maximum
3. Q = R√(C/L)
4. the resonant frequency is 1/(2π√(LC))
Show answer
Answer: A — the impedance seen by the source is at its maximum; C — Q = R√(C/L); D — the resonant frequency is 1/(2π√(LC))
In the parallel tank the inductive and capacitive susceptances cancel at ω₀ = 1/√(LC), leaving only G = 1/R, so the impedance is maximum (R) and Q = R/(ω₀L) = R√(C/L). A fixed-voltage source would therefore draw its minimum current, not its maximum — that is the series-resonance behaviour.