Given a 14 V source and drops of 5 V and 3 V, what remaining drop satisfies KVL?
A 8 V
B 5 V
C 3 V
D 6 V
Answer: D. 6 V. Kirchhoff’s voltage law requires the drops to sum to the source: 14 − 5 − 3 = 6 V.
With equal resistors across 12 V, what is an unloaded midpoint voltage?
A 3 V
B 12 V
C 6 V
D 9 V
Answer: C. 6 V. Equal divider resistors split the supply voltage equally.
When 2 A flows through a 10 Ω resistor, what power is dissipated?
A 20 W
B 200 W
C 40 W
D 5 W
Answer: C. 40 W. Using P = I²R gives 2² × 10 = 40 W.
At a node, 4 mA and 3 mA enter while 2 mA leaves. What other current leaves?
A 3 mA
B 5 mA
C 1 mA
D 7 mA
Answer: B. 5 mA. Kirchhoff’s current law gives 4 + 3 = 2 + I, so I = 5 mA.
With 4 µF capacitance and 5 V across it, what charge is stored?
A 25 µC
B 20 µC
C 9 µC
D 0.8 µC
Answer: B. 20 µC. Q = CV = 4 µF × 5 V = 20 µC.
R = 10 kΩ and C = 100 µF produce what RC time constant?
A 0.1 s
B 10 s
C 1 s
D 100 s
Answer: C. 1 s. The time constant is RC = 10,000 Ω × 100 µF = 1 s.
Using L = 2 H and R = 4 Ω, what is the RL time constant?
A 0.5 s
B 2 s
C 8 s
D 0.25 s
Answer: A. 0.5 s. An RL time constant is L/R = 2 H/4 Ω = 0.5 s.
At ordinary current, a silicon diode drop is roughly what?
A 5 V
B 0.3 V
C 0.7 V
D 1.0 V
Answer: C. 0.7 V. A silicon PN diode commonly has a forward drop near 0.7 V, depending on current and temperature.
For an ideal half-wave rectifier with a 12 V peak input, what average output results?
A 12 V
B 3.82 V
C 6 V
D 7.64 V
Answer: B. 3.82 V. The average half-wave output is Vm/π, which is 12/π ≈ 3.82 V.
In a full-wave rectifier, ripple frequency equals what?
A Twice supply frequency
B Supply frequency
C Half supply frequency
D Four times frequency
Answer: A. Twice supply frequency. Both input half-cycles produce output pulses, doubling the ripple frequency.
Ignoring diode drops, a bridge rectifier’s output peak is about what?
A Half input peak
B Input peak
C Twice input peak
D Zero volts
Answer: B. Input peak. The bridge redirects both half-cycles; ideally, the load sees approximately the input peak.
In an NPN transistor’s active region, how are its two junctions biased?
A Both junctions forward
B Both junctions reverse
C BE reverse, BC forward
D BE forward, BC reverse
Answer: D. BE forward, BC reverse. Active operation uses a forward-biased base-emitter junction and a reverse-biased base-collector junction.
For β = 100 and Ib = 20 µA, what collector current is estimated?
A 200 mA
B 0.2 mA
C 2 mA
D 20 mA
Answer: C. 2 mA. IC ≈ βIB = 100 × 20 µA = 2 mA.
When base current rises in a fixed-resistor common-emitter stage, what happens to collector voltage?
A It stays fixed
B It decreases
C It becomes negative
D It increases
Answer: B. It decreases. Higher base current raises collector current, increasing the collector-resistor drop and lowering collector voltage.
Which term names a transistor’s no-signal operating point?
A Cutoff point
B Ripple point
C Threshold point
D Q-point
Answer: D. Q-point. The quiescent or Q-point is the transistor’s DC operating point without an input signal.
Ideal MOSFET operation draws approximately what steady current at its gate?
A Zero
B High current
C One ampere
D Infinite current
Answer: A. Zero. The insulated gate draws essentially no steady-state current in the ideal model.
For Rf = 30 kΩ and Rin = 10 kΩ, what gain does an inverting op amp have?
A −4
B −3
C +3
D +4
Answer: B. −3. The inverting gain is −Rf/Rin = −30/10 = −3.
With Rf = 30 kΩ and Rg = 10 kΩ, what gain results in a non-inverting op amp?
A −4
B 3
C 4
D 10
Answer: C. 4. Its gain is 1 + Rf/Rg = 1 + 30/10 = 4.
Without feedback, an ideal op amp usually drives toward what?
A Supply rail
B Zero volts
C Virtual ground
D Midpoint voltage
Answer: A. Supply rail. Large open-loop gain drives the output to a supply rail for even a small input difference.
What cutoff results from R = 1 kΩ and C ≈ 0.159 µF?
A About 100 Hz
B About 1 MHz
C About 10 kHz
D About 1 kHz
Answer: D. About 1 kHz. Using fc = 1/(2πRC) gives approximately 1 kHz.
What does a diode’s reverse-recovery time measure?
A Time to stop reverse current
B Forward voltage rise
C Stored charge at rest
D Peak reverse voltage
Answer: A. Time to stop reverse current. Reverse-recovery time is the interval needed for a conducting diode to stop carrying current after reverse bias is applied.
Which circuit uses diodes to clip signal peaks?
A Clamper
B Integrator
C Oscillator
D Clipper
Answer: D. Clipper. A clipper limits portions of a waveform above or below selected levels.
Clamper circuits primarily shift which signal level?
A DC level
B Carrier amplitude
C Phase angle
D Frequency
Answer: A. DC level. A clamper adds a DC offset to shift a waveform’s level.
Given a 1:2 primary-to-secondary turns ratio, how does secondary voltage change?
A Reverses polarity
B Keeps voltage equal
C Halves secondary voltage
D Doubles secondary voltage
Answer: D. Doubles secondary voltage. For an ideal transformer, voltage ratio equals turns ratio, so twice as many secondary turns double the voltage.
How is a voltmeter connected across a component?
A In parallel
B In series
C Across source only
D In the ground lead
Answer: A. In parallel. A voltmeter measures the potential difference across a component, so it is connected in parallel.