Transistors & Logic Gates — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Transistors & Logic Gates MCQs with step-by-step solutions (20 questions). Part of Semiconductor Devices. Practise online on Prepizo — no login needed.
▶ Practise Transistors & Logic Gates online (free)
Questions with solutions
Q1 — Transistors & Logic Gates · easy · theory
A junction transistor consists of three regions called the:
A. Emitter, base and collector ✓ Correct
B. Primary, secondary and core
C. Source, gate and drain
D. Anode, cathode and grid
Solution: The emitter supplies carriers, the base controls them and the collector gathers them.
Q2 — Transistors & Logic Gates · medium · theory
The base region of a transistor is made thin and lightly doped so that:
A. Most of the injected carriers reach the collector rather than recombining ✓ Correct
B. The transistor cannot amplify
C. The collector current becomes zero
D. The base current is as large as possible
Solution: This is what makes $\alpha$ close to unity and gives the transistor its large current gain.
Q3 — Transistors & Logic Gates · medium · theory
In a common-emitter amplifier, the phase difference between the input and output signals is:
A. $270^\circ$
B. $90^\circ$
C. $180^\circ$ ✓ Correct
D. $0^\circ$
Solution: A rise in base current increases the collector current, which increases the drop across the load and lowers the output voltage.
Q4 — Transistors & Logic Gates · medium · theory
The universal logic gates are:
A. OR and NOT
B. XOR and AND
C. NAND and NOR ✓ Correct
D. AND and OR
Solution: Any logic function whatever can be built using only NAND gates, or only NOR gates.
Q5 — Transistors & Logic Gates · easy · theory
An AND gate gives an output of $1$ only when:
A. Its inputs differ
B. All its inputs are $0$
C. All its inputs are $1$ ✓ Correct
D. Any one input is $1$
Solution: Its Boolean expression is $Y = A \cdot B$.
Q6 — Transistors & Logic Gates · easy · theory
An OR gate gives an output of $1$ when:
A. Its inputs are equal
B. All its inputs are $0$
C. Any one or more of its inputs is $1$ ✓ Correct
D. All its inputs are $1$
Solution: Its Boolean expression is $Y = A + B$.
Q7 — Transistors & Logic Gates · easy · theory
A NOT gate performs the operation of:
A. Comparison of two inputs
B. Inversion of its single input ✓ Correct
C. Multiplication of two inputs
D. Addition of two inputs
Solution: Its Boolean expression is $Y = \bar{A}$.
Q8 — Transistors & Logic Gates · medium · theory
A transistor used in the cut-off and saturation regions only is working as:
A. A linear amplifier
B. An oscillator
C. A rectifier
D. A switch ✓ Correct
Solution: Between these two states it is either fully off or fully on, exactly like a mechanical switch.
Q9 — Transistors & Logic Gates · medium · numerical
The relation between the common-emitter and common-base current gains is:
A. $\alpha = \dfrac{\beta}{1 - \beta}$
B. $\beta = \dfrac{\alpha}{1 - \alpha}$ ✓ Correct
C. $\beta = \dfrac{1 + \alpha}{\alpha}$
D. $\alpha\beta = 1$
Solution: Since $\alpha$ is just less than unity, $\beta$ comes out in the tens or hundreds.
Q10 — Transistors & Logic Gates · medium · numerical
A transistor has a common-emitter current gain $\beta = 49$. Its common-base current gain $\alpha$ is:
A. $0.98$ ✓ Correct
B. $0.96$
C. $1.02$
D. $0.99$
Solution: $\alpha = \dfrac{\beta}{1 + \beta} = \dfrac{49}{50} = 0.98$.
Q11 — Transistors & Logic Gates · medium · numerical
A transistor has $\beta = 99$. Its value of $\alpha$ is:
A. $1.01$
B. $0.98$
C. $0.99$ ✓ Correct
D. $0.95$
Solution: $\alpha = \dfrac{99}{100} = 0.99$.
Q12 — Transistors & Logic Gates · medium · numerical
A transistor has $\alpha = 0.95$. Its value of $\beta$ is:
A. $20$
B. $19$ ✓ Correct
C. $95$
D. $0.95$
Solution: $\beta = \dfrac{\alpha}{1 - \alpha} = \dfrac{0.95}{0.05} = 19$.
Q13 — Transistors & Logic Gates · medium · numerical
A transistor with $\beta = 100$ has a base current of $20\,\mu\text{A}$. Its collector current is:
A. $0.2\text{ mA}$
B. $20\text{ mA}$
C. $2\text{ mA}$ ✓ Correct
D. $2\,\mu\text{A}$
Solution: $I_C = \beta I_B = 100 \times 20 \times 10^{-6} = 2\text{ mA}$.
Q14 — Transistors & Logic Gates · hard · numerical
A transistor has a base current of $20\,\mu\text{A}$ and a collector current of $2\text{ mA}$. Its emitter current is:
A. $20\,\mu\text{A}$
B. $1.98\text{ mA}$
C. $2\text{ mA}$
D. $2.02\text{ mA}$ ✓ Correct
Solution: $I_E = I_B + I_C = 0.02 + 2 = 2.02\text{ mA}$.
Q15 — Transistors & Logic Gates · hard · numerical
A transistor has a collector current of $1.98\text{ mA}$ and an emitter current of $2\text{ mA}$. Its value of $\alpha$ is:
A. $0.98$
B. $0.99$ ✓ Correct
C. $1.01$
D. $0.95$
Solution: $\alpha = \dfrac{I_C}{I_E} = \dfrac{1.98}{2} = 0.99$.
Q16 — Transistors & Logic Gates · medium · numerical
A NAND gate produces an output of $0$ only when:
A. All its inputs are $0$
B. All its inputs are $1$ ✓ Correct
C. Any input is $1$
D. Its inputs differ
Solution: $Y = \overline{A \cdot B}$, so the output is low only for the single input combination that makes the AND high.
Q17 — Transistors & Logic Gates · medium · numerical
The Boolean expression $Y = \overline{A + B}$ represents which gate?
A. NOR ✓ Correct
B. XOR
C. NAND
D. AND
Solution: It is the OR operation followed by inversion.
Q18 — Transistors & Logic Gates · medium · numerical
The decimal equivalent of the binary number $(1101)_2$ is:
A. $9$
B. $15$
C. $11$
D. $13$ ✓ Correct
Solution: $1(8) + 1(4) + 0(2) + 1(1) = 13$.
Q19 — Transistors & Logic Gates · medium · numerical
The decimal equivalent of the binary number $(1010)_2$ is:
A. $10$ ✓ Correct
B. $5$
C. $12$
D. $8$
Solution: $1(8) + 0(4) + 1(2) + 0(1) = 10$.
Q20 — Transistors & Logic Gates · hard · numerical
A transistor amplifier has $\beta = 100$, a collector load of $2\text{ k}\Omega$ and an input resistance of $1\text{ k}\Omega$. Its voltage gain is:
A. $200$ ✓ Correct
B. $2$
C. $100$
D. $50$
Solution: Voltage gain $= \beta \times \dfrac{R_C}{R_{in}} = 100 \times \dfrac{2000}{1000} = 200$.