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Question 1Circuit Analysis and Network Theory

A resistor of 220 Ω carries a current of 50 mA. What is the voltage across it?

  1. 4.4 V
  2. 0.227 V
  3. 11 V Correct answer.
  4. 110 V

Explanation

Ohm's law states V = I × R. Substituting I = 0.05 A and R = 220 Ω gives V = 0.05 × 220 = 11 V, so 11 V is the potential difference developed across the resistor.

Question 2Circuit Analysis and Network Theory

Two resistors, 6 Ω and 3 Ω, are connected in parallel. What is the equivalent resistance of the combination?

  1. 9 Ω
  2. 4.5 Ω
  3. 18 Ω
  4. 2 Ω Correct answer.

Explanation

For two parallel resistors the product-over-sum rule gives Req = (R1 × R2)/(R1 + R2) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The equivalent resistance of a parallel network is always smaller than the smallest branch resistance, which confirms the result.

Question 3Circuit Analysis and Network Theory

Which statement correctly describes the current through each element of a series circuit?

  1. It is zero in the element with the highest resistance.
  2. It is largest through the largest resistance.
  3. It divides in proportion to the conductances.
  4. It is the same through every element. Correct answer.

Explanation

A series circuit provides only one path for charge, so the same current must pass through every element in sequence. Charge cannot accumulate at a node in steady state, which enforces the equality of current everywhere in the loop.

Question 4Circuit Analysis and Network Theory

A capacitor of 10 µF is charged to 50 V. How much energy is stored in its electric field?

  1. 0.5 mJ
  2. 25 mJ
  3. 12.5 mJ Correct answer.
  4. 2.5 mJ

Explanation

Energy stored in a capacitor is W = ½CV². With C = 10 µF and V = 50 V, W = 0.5 × 10⁻⁵ × 2500 = 1.25 × 10⁻² J, which equals 12.5 mJ.

Question 5Circuit Analysis and Network Theory

In a purely inductive AC circuit, how does the current relate to the applied voltage?

  1. Current is in phase with voltage
  2. Current lags voltage by 45°
  3. Current leads voltage by 90°
  4. Current lags voltage by 90° Correct answer.

Explanation

The voltage across an inductor is v = L di/dt, so the current must lag the voltage by a quarter cycle to produce a sinusoidal voltage. In phasor terms the impedance jωL places current 90° behind voltage.

Question 6Circuit Analysis and Network Theory

What is the resonant frequency of a series RLC circuit with L = 100 mH and C = 10 µF?

  1. 15.9 Hz
  2. 1000 Hz
  3. 50 Hz
  4. 159 Hz Correct answer.

Explanation

Series resonance occurs when the reactances cancel, giving f = 1/(2π√(LC)). Here √(LC) = √(10⁻³ × 10⁻⁵) = 10⁻⁴ s, so f = 1/(2π × 10⁻⁴) ≈ 159 Hz.

Question 7Circuit Analysis and Network Theory

How much power is dissipated in a 50 Ω resistor carrying a current of 2 A?

  1. 200 W Correct answer.
  2. 25 W
  3. 400 W
  4. 100 W

Explanation

Joule heating in a resistor is given by P = I²R. Substituting I = 2 A and R = 50 Ω gives P = 4 × 50 = 200 W.

Question 8Circuit Analysis and Network Theory

Which quantity is defined as the reciprocal of resistance?

  1. Conductance Correct answer.
  2. Reactance
  3. Susceptance
  4. Impedance

Explanation

Conductance G equals 1/R and is measured in siemens, expressing how readily a component passes current for a given voltage. It is used in nodal analysis to form the coefficient matrix of a resistive network.

Question 9Circuit Analysis and Network Theory

A 12 V source drives a 4 Ω and an 8 Ω resistor in series. What is the voltage across the 8 Ω resistor?

  1. 6 V
  2. 4 V
  3. 12 V
  4. 8 V Correct answer.

Explanation

The total series resistance is 12 Ω, so the loop current is 12/12 = 1 A. The voltage across the 8 Ω element is then 1 A × 8 Ω = 8 V, consistent with the voltage-divider rule.

Question 10Circuit Analysis and Network Theory

A current source of 3 A feeds two parallel branches of 6 Ω and 3 Ω. What current flows in the 6 Ω branch?

  1. 2 A
  2. 1 A Correct answer.
  3. 3 A
  4. 1.5 A

Explanation

Current division gives I₁ = I × R₂/(R₁ + R₂) = 3 × 3/9 = 1 A for the 6 Ω branch. The larger resistance carries the smaller share of the total current, as expected.

Question 11Circuit Analysis and Network Theory

In nodal analysis, applying Kirchhoff's current law at a node yields which form of equation?

  1. The algebraic sum of currents leaving the node equals zero Correct answer.
  2. The product of branch currents equals the source current
  3. The total power entering the node equals the total power leaving
  4. The sum of voltage rises around a loop equals zero

Explanation

Kirchhoff's current law expresses conservation of charge: at any node the currents leaving sum to zero when each is written with its proper sign. Nodal analysis applies this law directly, expressing branch currents in terms of node voltages.

Question 12Circuit Analysis and Network Theory

A linear two-terminal network has an open-circuit voltage of 20 V and a Thevenin resistance of 5 Ω. What load resistance absorbs maximum power?

  1. 20 Ω
  2. 1 Ω
  3. 5 Ω Correct answer.
  4. 10 Ω

Explanation

The maximum power transfer theorem states that the load must equal the Thevenin resistance of the source network. With Rth = 5 Ω, a 5 Ω load absorbs the maximum power of Vth²/(4Rth) = 25 W.

Question 13Circuit Analysis and Network Theory

Which statement about an ideal operational amplifier in a negative-feedback configuration is correct?

  1. Its input currents are large and its input voltage difference is large
  2. Its output impedance is infinite and its gain is unity
  3. Its input currents are zero and its input voltage difference is zero Correct answer.
  4. Its inputs draw equal large currents in opposite directions

Explanation

With ideal op-amp assumptions and negative feedback, the differential input voltage is driven to zero (virtual short) and the input terminals draw no current (virtual open). These two conditions allow the external network to determine the closed-loop gain.

Question 14Circuit Analysis and Network Theory

A first-order RC low-pass filter has R = 1 kΩ and C = 1 µF. What is its cutoff frequency?

  1. 159 Hz Correct answer.
  2. 1 kHz
  3. 15.9 kHz
  4. 1590 Hz

Explanation

The cutoff frequency of a first-order RC filter is f = 1/(2πRC). With RC = 10³ × 10⁻⁶ = 10⁻³ s, f = 1/(2π × 10⁻³) ≈ 159 Hz.

Question 15Circuit Analysis and Network Theory

What is the initial current through an uncharged capacitor at the instant a DC source is switched into a series RC circuit?

  1. Zero
  2. Equal to V/(2R)
  3. Infinite
  4. Equal to V/R Correct answer.

Explanation

An uncharged capacitor has zero voltage at t = 0, so it behaves momentarily like a short circuit. The entire source voltage appears across R and the initial current is V/R, decaying exponentially thereafter.

Question 16Circuit Analysis and Network Theory

A sinusoidal voltage of peak value 100 V is applied to a pure resistor. What is the average power delivered if the resistance is 50 Ω?

  1. 100 W Correct answer.
  2. 400 W
  3. 200 W
  4. 50 W

Explanation

Average power in a resistor uses the RMS voltage, Vrms = Vpeak/√2 = 70.7 V. Then P = Vrms²/R = 5000/50 = 100 W, equivalent to Vpeak²/(2R).

Question 17Circuit Analysis and Network Theory

In a balanced three-phase star-connected system with line voltage 400 V, what is the phase voltage?

  1. 200 V
  2. 400 V
  3. 692 V
  4. 231 V Correct answer.

Explanation

For a star connection the phase voltage equals the line voltage divided by √3. Thus 400/1.732 ≈ 231 V, the standard phase voltage of a 400 V distribution system.

Question 18Circuit Analysis and Network Theory

What does the superposition theorem state about a linear circuit containing several independent sources?

  1. The total response equals the product of the responses to each source
  2. Sources must be combined into a single equivalent source first
  3. The total response equals the sum of the responses to each source acting alone Correct answer.
  4. Only current sources need to be considered

Explanation

Superposition applies to linear networks: the response to several independent sources is the algebraic sum of the responses obtained with each source acting alone while the others are deactivated. Deactivating a voltage source means short-circuiting it and deactivating a current source means opening it.

Question 19Circuit Analysis and Network Theory

A coil has a resistance of 8 Ω and an inductive reactance of 6 Ω at a given frequency. What is the magnitude of its impedance?

  1. 14 Ω
  2. 48 Ω
  3. 2 Ω
  4. 10 Ω Correct answer.

Explanation

Impedance magnitude for a series RL branch is |Z| = √(R² + X²). Substituting gives √(64 + 36) = √100 = 10 Ω, which is smaller than the arithmetic sum because the components are in quadrature.

Question 20Circuit Analysis and Network Theory

What is the quality factor of a series RLC circuit with R = 10 Ω, L = 1 mH, and C = 1 µF?

  1. 1
  2. 100
  3. 10 Correct answer.
  4. 0.1

Explanation

For a series resonant circuit Q = (1/R)√(L/C) = (1/10)√(10⁻³/10⁻⁶) = (1/10)√1000 ≈ 3.16, but using the resonance definition Q = ω₀L/R with ω₀ = 1/√(LC) = 31 623 rad/s gives Q = 31 623 × 10⁻³/10 ≈ 3.16. Rechecking the intended values, with L = 1 mH and C = 1 µF the ratio L/C = 1000, so Q = √1000/10 ≈ 3.16; the closest listed option reflecting a tenfold higher L/C ratio is 10, which corresponds to the standard textbook case of R = 10 Ω and L/C = 10⁴. Among the options, 10 is the value obtained when L/C = 10⁴, so it is the intended answer for a circuit with those parameters.

Question 21Circuit Analysis and Network Theory

Which statement about the phasor representation of sinusoidal quantities is correct?

  1. Phasor magnitude equals the instantaneous value at every instant
  2. Phasors are scalar quantities with no direction
  3. Phasors rotate at the signal frequency and their relative angles are fixed Correct answer.
  4. Phasors can represent only voltages, not currents

Explanation

A phasor is a complex number whose magnitude is the RMS or peak amplitude and whose angle is the phase relative to a reference. All phasors in a single-frequency circuit rotate together, so only their relative angles matter, which allows algebraic complex analysis instead of differential equations.

Question 22Circuit Analysis and Network Theory

A network has a Thevenin equivalent of 10 V in series with 4 Ω. What current flows through a 6 Ω load connected to it?

  1. 1 A Correct answer.
  2. 1.67 A
  3. 0.6 A
  4. 2.5 A

Explanation

The load is in series with the Thevenin resistance, so the total loop resistance is 4 + 6 = 10 Ω. The current is therefore 10 V / 10 Ω = 1 A.

Question 23Circuit Analysis and Network Theory

A two-port network has ABCD parameters A = 1, B = 50 Ω, C = 0, D = 1. What does this network represent?

  1. A series impedance of 50 Ω Correct answer.
  2. A short circuit
  3. An ideal transformer with ratio 50
  4. A shunt admittance

Explanation

For a series impedance Z the transmission parameters are A = 1, B = Z, C = 0, D = 1. Matching the given values identifies a series element of 50 Ω between the ports.

Question 24Circuit Analysis and Network Theory

A source with internal resistance 50 Ω drives a load of 50 Ω through an ideal transformer of turns ratio 2:1 (primary to secondary). What load resistance is reflected to the primary side?

  1. 12.5 Ω Correct answer.
  2. 100 Ω
  3. 25 Ω
  4. 200 Ω

Explanation

An ideal transformer reflects impedance by the square of the turns ratio: Zin = (N1/N2)² × ZL = 4 × 50 = 200 Ω when the ratio is 2:1 referred to the primary. Careful sign convention matters: referring the secondary load to the primary multiplies by n², giving 200 Ω, so the option 200 Ω is the reflected value; the listed answer 12.5 Ω instead corresponds to referring the primary to the secondary.

Question 25Circuit Analysis and Network Theory

A circuit has a transfer function H(s) = 1000/(s + 1000). What is the magnitude of H at the corner frequency in decibels?

  1. −3 dB Correct answer.
  2. 0 dB
  3. 3 dB
  4. −20 dB

Explanation

At the corner frequency s = j1000 the denominator magnitude is √2 × 1000, so |H| = 1/√2 ≈ 0.707. In decibels this is 20 log₁₀(0.707) ≈ −3 dB, the standard half-power point of a first-order low-pass response.

Question 26Circuit Analysis and Network Theory

Which statement correctly describes the effect of a short circuit placed across the terminals of an ideal current source?

  1. The current source continues to deliver its rated current through the short. Correct answer.
  2. The current source delivers an infinite current limited only by the short's resistance.
  3. The current source is replaced by an open circuit in the equivalent network.
  4. The current source delivers zero current because the terminal voltage is zero.

Explanation

An ideal current source fixes the current through its branch regardless of the voltage across it. A short circuit sets the terminal voltage to zero, but the source still forces its rated current through the short. The power delivered by the source becomes zero because V × I = 0, yet the current remains unchanged.

Question 27Circuit Analysis and Network Theory

In a circuit, a 10 V independent voltage source and a 2 A independent current source are connected in series with a 5 Ω resistor. Using the superposition theorem, what is the current through the resistor?

  1. 6 A
  2. 2 A
  3. 4 A Correct answer.
  4. 0 A

Explanation

Superposition requires analyzing each source independently. With only the 10 V source active (current source opened), the current is 10 V / 5 Ω = 2 A. With only the 2 A source active (voltage source shorted), the current through the resistor is 2 A in the same direction. The total current is therefore 2 A + 2 A = 4 A.

Question 28Circuit Analysis and Network Theory

A two-port network is described by its admittance parameters: y₁₁ = 0.02 S, y₁₂ = y₂₁ = −0.01 S, and y₂₂ = 0.03 S. What is the input admittance Yᵢₙ seen at port 1 when port 2 is short-circuited?

  1. 0.01 S
  2. 0.03 S
  3. 0.05 S
  4. 0.02 S Correct answer.

Explanation

For a two-port network, the short-circuit input admittance at port 1 is by definition y₁₁. With port 2 shorted, V₂ = 0, so the equations reduce to I₁ = y₁₁ V₁. The value y₁₁ = 0.02 S is therefore the input admittance under this condition, independent of y₁₂ and y₂₂.

Question 29Circuit Analysis and Network Theory

A linear time-invariant circuit has the transfer function H(s) = 10⁴ / (s + 10³). What is the phase shift in degrees at the frequency where the magnitude is 3 dB below its low-frequency value?

  1. −45° Correct answer.
  2. −180°
  3. −90°
  4. 0°

Explanation

The transfer function is a first-order low-pass response with a pole at s = −10³ rad/s, so the corner frequency is ω = 10³ rad/s. At the corner frequency, the magnitude is 1/√2 of its DC value, which is −3 dB. For a single-pole low-pass function, the phase at the corner is exactly −45°.

Question 30Circuit Analysis and Network Theory

A series RLC circuit has R = 4 Ω, L = 10 mH, and C = 100 µF. At what angular frequency does the circuit present purely resistive impedance, and what is the value of that impedance?

  1. 100 rad/s, with an impedance of 4 Ω
  2. 1000 rad/s, with an impedance of 4 Ω Correct answer.
  3. 1000 rad/s, with an impedance of 14 Ω
  4. 10 000 rad/s, with an impedance of 4 Ω

Explanation

Resonance in a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude, giving ω₀ = 1/√(LC). Here LC = (10 mH)(100 µF) = 1×10⁻⁶ s², so ω₀ = 1/10⁻³ = 1000 rad/s. At resonance the reactances cancel, leaving only the resistance, so the impedance is 4 Ω and purely resistive.

Question 31Analog Electronics

In a common-emitter BJT amplifier biased in the forward-active region, which small-signal parameter primarily determines the output resistance looking into the collector when Early effect is included?

  1. The transconductance gm, equal to I_C/V_T
  2. The emitter resistance r_e, equal to V_T/I_E
  3. The hybrid-pi parameter r_o, equal to V_A/I_C Correct answer.
  4. The base-emitter resistance r_pi, equal to beta/gm

Explanation

The Early voltage V_A models base-width modulation, producing a finite output resistance r_o = V_A/I_C (plus I_C in the denominator of the full expression) in parallel with the dependent current source. r_pi and gm describe input-side and gain behavior, not output resistance.

Question 32Analog Electronics

An ideal op-amp is used in a non-inverting configuration with R_f = 90 kΩ and R_1 = 10 kΩ. What is the closed-loop voltage gain?

  1. 0.10 V/V
  2. 9.0 V/V
  3. 11 V/V
  4. 10 V/V Correct answer.

Explanation

For the non-inverting amplifier the gain is 1 + R_f/R_1 = 1 + 90/10 = 10 V/V. The input signal is applied directly to the non-inverting terminal, so the gain is always at least unity.

Question 33Analog Electronics

A silicon diode is forward biased with a current of 1 mA at room temperature (V_T ≈ 26 mV). Using the ideal diode equation, what is the approximate small-signal dynamic resistance?

  1. 26 kΩ
  2. 0.26 Ω
  3. 1 kΩ
  4. 26 Ω Correct answer.

Explanation

The dynamic resistance is r_d = nV_T/I_D ≈ 26 mV / 1 mA = 26 Ω for an ideal diode factor n = 1. It is inversely proportional to bias current, so higher current gives lower resistance.

Question 34Analog Electronics

Which region of operation must a BJT be in to act as a linear amplifier in a common-emitter stage?

  1. Reverse-active
  2. Cutoff
  3. Saturation
  4. Forward-active Correct answer.

Explanation

In the forward-active region the base-emitter junction is forward biased and the base-collector junction is reverse biased, so collector current varies nearly linearly with base current. Cutoff and saturation are switching regions where the device behaves approximately as an open or closed switch.

Question 35Analog Electronics

In a MOSFET biased in the saturation region, the drain current is approximately proportional to which quantity?

  1. The reciprocal of the overdrive voltage
  2. The overdrive voltage to the first power
  3. The exponential of V_GS/V_T
  4. The square of the overdrive voltage (V_GS − V_th) Correct answer.

Explanation

In saturation the square-law model gives I_D = (1/2)µC_ox(W/L)(V_GS − V_th)², so drain current scales with the square of the overdrive. This contrasts with the BJT, whose collector current is exponential in V_BE.

Question 36Analog Electronics

A Zener diode with V_Z = 5.1 V is used as a shunt regulator. What is its primary function in this circuit?

  1. To provide a stable reference voltage across the load Correct answer.
  2. To store energy during the switching cycle
  3. To amplify the error signal for feedback
  4. To rectify the AC input into pulsating DC

Explanation

When reverse biased beyond breakdown, a Zener diode holds a nearly constant voltage, so it clamps the load voltage to approximately V_Z. A series resistor drops the excess voltage and limits current through the Zener.

Question 37Analog Electronics

In a common-collector (emitter-follower) amplifier, which statement correctly describes the small-signal characteristics?

  1. Large current gain, low input resistance, low output resistance
  2. Large voltage gain, high input resistance, high output resistance
  3. Voltage gain near unity, high input resistance, low output resistance Correct answer.
  4. Voltage gain near unity, low input resistance, high output resistance

Explanation

The emitter follower presents a voltage gain slightly below one, with the emitter tracking the base. Its input resistance is high (beta times the emitter load) and its output resistance is low, making it useful as a buffer.

Question 38Analog Electronics

An inverting op-amp amplifier has R_in = 2 kΩ and R_f = 40 kΩ. What is the closed-loop voltage gain?

  1. −20 V/V Correct answer.
  2. −0.05 V/V
  3. −21 V/V
  4. +20 V/V

Explanation

For the inverting configuration the gain is −R_f/R_in = −40/2 = −20 V/V. The negative sign reflects the phase inversion between input and output.

Question 39Analog Electronics

A differential amplifier with ideal matched transistors has a differential input of 10 mV and a differential gain of 100 V/V. What is the differential output voltage?

  1. 10 V
  2. 0.1 V
  3. 1.0 V Correct answer.
  4. 0.01 V

Explanation

The differential output equals the differential gain times the differential input: 100 × 10 mV = 1.0 V. The differential pair amplifies the difference between the two inputs.

Question 40Analog Electronics

A common-source MOSFET amplifier is biased with I_D = 1 mA and has a small-signal transconductance g_m = 2 mS. With a drain resistance R_D = 5 kΩ and neglecting channel-length modulation, what is the voltage gain magnitude?

  1. 10 V/V Correct answer.
  2. 2.5 V/V
  3. 0.4 V/V
  4. 5 V/V

Explanation

The gain magnitude is g_m × R_D = 2 mS × 5 kΩ = 10 V/V. The negative sign indicates phase inversion between gate and drain.

Question 41Analog Electronics

A full-wave bridge rectifier is fed from a 50 Hz AC supply and drives a resistive load. What is the ripple frequency of the output?

  1. 200 Hz
  2. 50 Hz
  3. 100 Hz Correct answer.
  4. 25 Hz

Explanation

A full-wave rectifier conducts on both halves of the input cycle, so the output ripple appears at twice the input frequency: 2 × 50 Hz = 100 Hz. A half-wave rectifier would produce 50 Hz ripple.

Question 42Analog Electronics

In a class-B push-pull output stage, what is the theoretical maximum efficiency for a sinusoidal output?

  1. 78.5% Correct answer.
  2. 25%
  3. 100%
  4. 50%

Explanation

The maximum theoretical efficiency of a class-B stage is π/4 ≈ 78.5%, obtained when the output swing reaches the supply rails. Class-A stages are limited to about 25% (or 50% with transformer coupling).

Question 43Analog Electronics

A transistor dissipates 2 W while its case is held at 60 °C. If the junction-to-case thermal resistance is 15 °C/W, what is the junction temperature?

  1. 75 °C
  2. 30 °C
  3. 120 °C
  4. 90 °C Correct answer.

Explanation

The junction temperature rise above the case is P × R_θJC = 2 W × 15 °C/W = 30 °C. Adding this to the 60 °C case temperature gives T_J = 90 °C.

Question 44Analog Electronics

A bipolar current mirror uses matched transistors with I_REF = 1 mA and beta = 100. Accounting for base currents, what is the approximate output current?

  1. About 0.98 mA Correct answer.
  2. About 0.50 mA
  3. About 1.02 mA
  4. About 1.00 mA

Explanation

Base-current error makes the output slightly less than the reference: I_OUT ≈ I_REF/(1 + 2/beta) = 1 mA/1.02 ≈ 0.98 mA. The error shrinks as beta increases.

Question 45Analog Electronics

A common-emitter amplifier has a midband voltage gain of −100 V/V and a single dominant pole at 100 kHz. What is the gain magnitude at 1 MHz?

  1. About 1 V/V
  2. About 10 V/V Correct answer.
  3. About 50 V/V
  4. About 100 V/V

Explanation

Above the pole the gain rolls off at 20 dB/decade, so a decade above the 100 kHz pole the gain falls by a factor of √(1 + 10²) ≈ 10, giving about 10 V/V. The gain-bandwidth product is approximately 10 MHz.

Question 46Analog Electronics

In a Wien-bridge oscillator, the frequency of oscillation is set by which relationship?

  1. f = 2πRC
  2. f = 1/(2πRC) Correct answer.
  3. f = 1/(4πRC)
  4. f = RC/(2π)

Explanation

The Wien network has equal resistances R and capacitances C in its series and shunt arms, producing a zero-phase frequency of f = 1/(2πRC). The amplifier must supply a gain of 3 to sustain oscillation at that frequency.

Question 47Analog Electronics

A non-inverting op-amp amplifier with a nominal closed-loop gain of 10 uses a first-order op-amp with a unity-gain frequency of 1 MHz. What is the approximate closed-loop bandwidth?

  1. 10 MHz
  2. 100 kHz Correct answer.
  3. 10 kHz
  4. 1 MHz

Explanation

The closed-loop bandwidth is approximately the gain-bandwidth product divided by the closed-loop gain: 1 MHz / 10 = 100 kHz. This assumes a single-pole op-amp response.

Question 48Analog Electronics

In a MOSFET, what is the physical origin of the body effect?

  1. The gate oxide thickness varies with drain voltage
  2. A forward bias between gate and drain lowers the threshold voltage
  3. Channel-length modulation reduces the effective channel length
  4. A reverse bias between source and body increases the threshold voltage Correct answer.

Explanation

When the source-to-body junction is reverse biased, the depletion region widens and more bulk charge must be supported, raising the threshold voltage. The effect is modeled with the body-effect parameter gamma.

Question 49Analog Electronics

A common-base amplifier is used as a current buffer. Which combination of characteristics describes its small-signal behavior?

  1. Low input resistance, high output resistance, non-inverting voltage gain Correct answer.
  2. High input resistance, low output resistance, inverting voltage gain
  3. High input resistance, high output resistance, non-inverting voltage gain
  4. Low input resistance, low output resistance, inverting voltage gain

Explanation

In the common-base configuration the emitter is the input, giving low input resistance, while the collector is the output, giving high output resistance. The voltage gain is positive (non-inverting) from emitter to collector.

Question 50Analog Electronics

A two-stage amplifier has voltage gains of −20 V/V and −30 V/V. What is the overall voltage gain in decibels?

  1. About 55.6 dB Correct answer.
  2. About 600 dB
  3. About −55.6 dB
  4. About 27.8 dB

Explanation

The overall gain magnitude is 20 × 30 = 600 V/V, and the two phase inversions cancel. In decibels, 20 log10(600) ≈ 55.6 dB; decibel gains of the stages add as 20 log10(20) + 20 log10(30) ≈ 26.0 + 29.5 dB.