With a slew rate of 0.5 V/µs, how long does a 5 V output change take?
A 5 µs
B 10 µs
C 2.5 µs
D 25 µs
Answer: B. 10 µs. The transition time is voltage change divided by slew rate: 5 V/(0.5 V/µs) = 10 µs.
Using an 18 V source with 2 kΩ and 4 kΩ in series, what voltage appears across 4 kΩ?
A 6 V
B 9 V
C 12 V
D 18 V
Answer: C. 12 V. The 3 mA series current gives V = IR = 3 mA × 4 kΩ = 12 V.
For a 12 V source with 2 kΩ series resistance, what Norton current results?
A 24 mA
B 12 mA
C 3 mA
D 6 mA
Answer: D. 6 mA. Norton current is the short-circuit current, V/R = 12 V/2 kΩ = 6 mA.
Shorting the ideal 12 V source leaves what output resistance when its series resistor is 2 kΩ?
A 1 kΩ
B 6 kΩ
C 2 kΩ
D 4 kΩ
Answer: C. 2 kΩ. Deactivate the ideal voltage source by shorting it; the output sees the 2 kΩ resistor.
Given R1 = 1 kΩ, R2 = 2 kΩ, and a 2 kΩ load across R2 on a 9 V divider, what is output?
A 6 V
B 9 V
C 4.5 V
D 3 V
Answer: C. 4.5 V. R2 in parallel with the load is 1 kΩ, giving 9 × 1/(1 + 1) = 4.5 V.
With VCE = 5 V and IC = 20 mA, what collector power is dissipated?
A 25 mW
B 200 mW
C 50 mW
D 100 mW
Answer: D. 100 mW. Transistor collector power is VCE × IC = 5 V × 20 mA = 100 mW.
After one RC time constant, charging reaches about what fraction?
A 50 percent
B 63.2 percent
C 95 percent
D 36.8 percent
Answer: B. 63.2 percent. A charging capacitor reaches 1 − e⁻¹, about 63.2% of its final value after one time constant.
After five RC time constants, charging is approximately what?
A About 99 percent
B About 86 percent
C About 63 percent
D About 50 percent
Answer: A. About 99 percent. At 5τ, the capacitor has reached 1 − e⁻⁵, approximately 99.3% of its final voltage.
Doubling R in a fixed-C RC filter does what to cutoff?
A Halves it
B Doubles it
C Leaves it unchanged
D Makes it zero
Answer: A. Halves it. Since fc = 1/(2πRC), doubling R halves the cutoff frequency.
At low frequencies, what does high-pass output approach?
A Infinite output
B Input peak
C Zero
D Supply voltage
Answer: C. Zero. At very low frequency, the series capacitor blocks the signal, so output approaches zero.
If the emitter is grounded and the base is fed from 10 V through 470 kΩ, with β = 100 and VBE = 0.7 V, what is IC?
A About 200 mA
B About 2 mA
C About 20 mA
D About 0.2 mA
Answer: B. About 2 mA. IB ≈ (10 − 0.7)/470 kΩ ≈ 19.8 µA, so βIB ≈ 1.98 mA.
Why does an emitter resistor help reduce BJT thermal runaway?
A Positive feedback
B Higher base current
C Lower supply voltage
D Negative feedback
Answer: D. Negative feedback. Rising collector current raises emitter voltage, reducing base-emitter drive and counteracting further current increase.
Why is a common-emitter amplifier’s Q-point often chosen near mid-load line?
A To allow symmetric signal swing
B To eliminate collector load
C To minimize base current
D To force cutoff continuously
Answer: A. To allow symmetric signal swing. A near-middle operating point allows a larger undistorted output swing before reaching cutoff or saturation.
What voltage gain is typical of an emitter follower?
A About zero
B About unity
C About ten
D Infinite gain
Answer: B. About unity. A common-collector emitter follower has voltage gain close to one.
Enhancement NMOS conduction begins once VGS exceeds what?
A Source resistance
B Gate current
C Drain voltage
D Threshold voltage
Answer: D. Threshold voltage. The channel forms when gate-to-source voltage exceeds the threshold voltage.
Saturation begins when which voltage condition holds for an enhancement NMOS?
A VDS ≥ VGS−Vth
B VDS = 0
C VGS ≤ Vth
D VDS ≤ VGS−Vth
Answer: A. VDS ≥ VGS−Vth. For the long-channel enhancement NMOS model, saturation begins when VDS is at least VGS − Vth.
Which ideal op-amp parameter minimizes output impedance?
A Zero output resistance
B Infinite output resistance
C Zero input resistance
D Low open-loop gain
Answer: A. Zero output resistance. An ideal op amp has zero output resistance, so its output voltage is unaffected by load current.
What makes a real op amp’s closed-loop gain fall at high frequencies?
A Infinite input resistance
B Ideal negative feedback
C Finite gain-bandwidth product
D Zero output resistance
Answer: C. Finite gain-bandwidth product. A real op amp has finite gain-bandwidth product, so its closed-loop gain decreases as signal frequency rises.
For an ideal op-amp integrator, which component occupies the feedback path?
A Transformer
B Zener diode
C Capacitor
D Inductor
Answer: C. Capacitor. An op-amp integrator places a capacitor in its feedback path, commonly with an input resistor.
Which signal component should an ideal differential amplifier reject?
A Common-mode signal
B Differential signal
C Output noise
D Input bias current
Answer: A. Common-mode signal. An ideal differential amplifier responds to the input difference and rejects equal common-mode signals.
During both input half-cycles, a bridge rectifier keeps load polarity how?
A Opposite directions
B Same direction
C Zero current
D Alternating voltage
Answer: B. Same direction. The diode bridge routes each half-cycle so current through the load remains in one direction.
Increasing filter capacitance, with load and frequency fixed, does what to ripple?
A Increases ripple
B Doubles frequency
C Stops charging
D Reduces ripple
Answer: D. Reduces ripple. A larger capacitor stores more charge between peaks, reducing output ripple.
How many diodes conduct during each bridge-rectifier half-cycle?
A Four
B Two
C One
D Three
Answer: B. Two. Two bridge diodes conduct in each half-cycle, routing current through the load in the same direction.
For a BJT, how is α defined?
A Ie divided by Ic
B Ic divided by Ib
C Ib divided by Ic
D Ic divided by Ie
Answer: D. Ic divided by Ie. The common-base current gain is α = IC/IE.
Two cascaded common-emitter stages yield what overall phase?
A In phase
B 180 degrees apart
C 270 degrees apart
D 90 degrees apart
Answer: A. In phase. Each common-emitter stage inverts; two inversions restore the original phase.