Q.A sealed box has three terminals A, B and C. Inside, two identical germanium diodes (each with a forward threshold of about $0.7\,\text{V}$) and one $1000\,\Omega$ resistor are connected in some arrangement among the terminals. Using a d.c. source, an ammeter and a voltmeter, the current-voltage ($I$-$V$) characteristic is measured between each pair of terminals, with the third terminal left unconnected. The results are:

Showing the 12 most recent of 23 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If A, B and C represent the power gain, base resistance and collector resistance respectively of a transistor connected in common emitter configuration, then the common emitter current amplification factor is (A) $\dfrac{AB}{C}$ (B) $\dfrac{AC}{B}$ (C) $\sqrt{\dfrac{AC}{B}}$ (D) $\sqrt{\dfrac{AB}{C}}$

   ›Reveal solutionSolution

   For a common-emitter transistor, power gain $A = \beta^2\,\dfrac{R_C}{R_B} = \beta^2\dfrac{C}{B}$, so $\beta = \sqrt{\dfrac{AB}{C}}$.

   In a common-emitter amplifier the power gain equals the current gain times the voltage gain:

   $A = \beta \times A_V = \beta \times \left(\beta\,\frac{R_C}{R_B}\right) = \beta^2\,\frac{R_C}{R_B}$

   With base resistance $B = R_B$ and collector resistance $C = R_C$: …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The incorrect statement about a Zener diode is (A) It is heavily doped (B) The depletion region formed in it is very thick (C) The electric field at its junction is extremely high (D) In general, in a circuit it is reverse biased

   ›Reveal solutionSolution

   A Zener diode is a heavily doped, reverse-biased device with a very thin depletion region and an extremely high junction field. The false statement is the one that claims the depletion region is thick.

   The key to this question lies in understanding why a Zener diode behaves differently from an ordinary diode. An ordinary p-n junction breaks down (conducts heavily in reverse bias) only at a high voltage, often destructively. A Zener diode is designed to break down at a precisely controlled, low reverse voltage and to survive that breakdown — this is its normal operating mode.

   That controlled, low-voltage breakdown is achieved by heavy doping. When both the p and n sides are doped very heavily, the depletion region becomes extremely narrow. Because the same built-in potential now exists across a much smaller distance, the electric field inside the depletion region becomes enormous — easily reaching millions of volts per meter. This intense field is what pulls electrons out of their covalent bonds (Zener breakdown) at a low applied voltage.

   Now let's examine each statement.

   1. Statement (A): "It is heavily doped"

      This is true. Heavy doping is the defining feature that gives the Zener diode its characteristic low breakdown voltage. Without it, the depletion region would be wider and the breakdown voltage would be much higher.

   2. Statement (B): "The depletion region formed in it is very thick"

      This is false. Heavy doping means there are more charge carriers available to neutralize the fixed ions near the junction. As a result, the depletion region is actually very thin — typically on the order of tens of nanometres. A thick depletion region is a property of a lightly doped diode, not a Zener diode. …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.An electron in n-region of a p-n junction diode moves towards the junction with a speed of $5 \times 10^5$ ms$^{-1}$ and after penetration through the barrier, the electron enters the p-region with a speed of $3 \times 10^5$ ms$^{-1}$. The barrier potential of the p-n junction is (Mass of electron $= 9 \times 10^{-31}$ kg and charge of electron $= 1.6 \times 10^{-19}$ C) (A) 0.5 V (B) 0.45 V (C) 0.7 V (D) 0.65 V

   ›Reveal solutionSolution

   The barrier potential is found by equating the loss in kinetic energy of the electron to the gain in electrical potential energy. The result is 0.45 V, so option (B) is correct.

   The key idea here is energy conservation. As the electron moves from the n-region to the p-region across the depletion layer, it must climb over the built-in potential barrier. This barrier opposes the motion of electrons from n to p, so the electron loses kinetic energy and gains electrical potential energy. The loss in kinetic energy equals the charge of the electron times the barrier potential.

   1. Identify the energy change The electron’s speed drops from $v_i = 5 \times 10^5$ m/s to $v_f = 3 \times 10^5$ m/s. Its kinetic energy decreases by

   $\Delta K = \frac{1}{2} m v_i^2 - \frac{1}{2} m v_f^2 = \frac{1}{2} m (v_i^2 - v_f^2).$

   2. Relate to electrical potential energy The electron gains electrical potential energy $e V_b$ (where $V_b$ is the barrier potential) because it moves against the electric field of the junction. By conservation of energy,

   $\frac{1}{2} m (v_i^2 - v_f^2) = e V_b.$

   3. Plug in the numbers Mass $m = 9 \times 10^{-31}$ kg, charge $e = 1.6 \times 10^{-19}$ C.

   $v_i^2 = (5 \times 10^5)^2 = 25 \times 10^{10} = 2.5 \times 10^{11},$

   $v_f^2 = (3 \times 10^5)^2 = 9 \times 10^{10} = 0.9 \times 10^{11},$

   $v_i^2 - v_f^2 = (2.5 - 0.9) \times 10^{11} = 1.6 \times 10^{11}.$

   Then

   $$$$

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4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.In common emitter amplifier, when the base to emitter voltage changes by 20 mV, the change in collector current is 3 mA. If the output resistance is $4~\text{k}\Omega$, then the voltage gain of the amplifier is (A) 300 (B) 600 (C) 400 (D) 800

   ›Reveal solutionSolution

   Voltage gain in a CE amplifier is the product of transconductance and output resistance. Here $g_m = \frac{\Delta I_C}{\Delta V_{BE}} = 150~\text{mS}$ and $R_o = 4~\text{k}\Omega$, so $A_V = 150 \times 4 = 600$.

   The key idea is that in a common emitter amplifier, the voltage gain is determined by how effectively a small change in base-emitter voltage controls the collector current, multiplied by the load it drives. The transconductance $g_m$ captures this control — it’s the ratio of change in collector current to change in base-emitter voltage. Once you have $g_m$, the voltage gain is simply $A_V = g_m R_o$, where $R_o$ is the output resistance (the load seen by the collector current).

   Let’s work it through.

   1. Find the transconductance $g_m$. The definition is $g_m = \frac{\Delta I_C}{\Delta V_{BE}}$. Given $\Delta V_{BE} = 20~\text{mV} = 0.02~\text{V}$ and $\Delta I_C = 3~\text{mA} = 0.003~\text{A}$,

   $g_m = \frac{0.003}{0.02} = 0.15~\text{S} = 150~\text{mS}.$

   2. Identify the output resistance.

      The problem states the output resistance is $4~\text{k}\Omega = 4000~\Omega$. In a CE amplifier, this is the resistance seen by the collector current — typically $R_C$ or $R_L$ depending on context. Here it’s given directly.

   3. Compute the voltage gain.

      For a common emitter amplifier, the voltage gain (magnitude) is

   $A_V = g_m R_o.$

   Substituting: …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.$10^{10}$ electrons enter the emitter of a junction transistor in a time of $0.4~\mu\text{s}$. If $5\%$ of the electrons are lost in the base, then the collector current is (A) $3.0~\text{mA}$ (B) $3.2~\text{mA}$ (C) $3.6~\text{mA}$ (D) $3.8~\text{mA}$

   ›Reveal solutionSolution

   The collector current is found from the number of electrons reaching the collector per second, accounting for the 5% loss in the base. The result is 3.8 mA, so option (D) is correct.

   Concept & Intuition

   In a bipolar junction transistor, electrons injected from the emitter into the base are mostly swept across to the collector. A small fraction recombine in the base (here 5%), so the collector current is 95% of the emitter current. The emitter current itself is the charge of the electrons entering per unit time. We just need to compute that charge flow and then apply the loss factor.

   Step-by-step solution

   1. Find the total charge entering the emitter.

      Each electron carries charge $e = 1.6 \times 10^{-19}~\text{C}$.

      Number of electrons = $10^{10}$.

      Total charge $Q = 10^{10} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-9}~\text{C}$.

   2. Compute the emitter current.

      Time interval $\Delta t = 0.4~\mu\text{s} = 0.4 \times 10^{-6}~\text{s}$. …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.An electron in n-region of a p-n junction moves towards the junction with a speed of $5 \times 10^5 \ \mathrm{ms}^{-1}$. If the barrier potential of the junction is $0.45 \ \mathrm{V}$, then the speed with which the electron enters the p-region after penetration through the barrier is (Charge of the electron $= 1.6 \times 10^{-19} \ \mathrm{C}$ and mass of the electron $= 9 \times 10^{-31} \ \mathrm{kg}$) (A) $3 \times 10^5 \ \mathrm{ms}^{-1}$ (B) $5 \times 10^5 \ \mathrm{ms}^{-1}$ (C) $4 \times 10^5 \ \mathrm{ms}^{-1}$ (D) $6 \times 10^5 \ \mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The electron enters the p-region with speed $3 \times 10^5\,\text{m/s}$ — option (A).

   Given: $v_i = 5\times 10^5\,\text{m/s}$, barrier potential $V = 0.45\,\text{V}$, $e = 1.6\times 10^{-19}\,\text{C}$, $m = 9\times 10^{-31}\,\text{kg}$.

   An electron is a majority carrier in the n-region; the built-in barrier decelerates it as it crosses to the p-region, so it loses kinetic energy $eV$.

   Energy conservation:

   $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 - eV.$

   Initial KE:

   $\frac{1}{2}mv_i^2 = \frac{1}{2}(9\times10^{-31})(5\times10^5)^2 = 1.125\times10^{-19}\,\text{J}.$

   Energy lost: …

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7. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.If the breakdown voltage of a Zener diode is 7.2 V and the maximum power dissipated is 216 mW, then the maximum current through the Zener diode is (A) 30 mA (B) 4.16 mA (C) 3.33 mA (D) 2.4 mA

   ›Reveal solutionSolution

   The maximum current through a Zener diode is found by dividing its maximum power dissipation by its Zener breakdown voltage. For the given values, the maximum current is $\boxed{\text{30 mA}}$.

   Concept and Intuition

   A Zener diode is a special type of diode designed to operate in the reverse breakdown region. When reverse biased, it maintains a nearly constant voltage across its terminals, known as the Zener voltage ($V_Z$), even as the current through it varies over a wide range. This property makes Zener diodes ideal for voltage regulation.

   Every electronic component has a limit to how much power it can dissipate without being damaged. For a Zener diode, this maximum power dissipation ($P_{max}$) is specified by the manufacturer. Power dissipation in any component is given by the product of the voltage across it and the current flowing through it ($P = V \times I$).

   In the Zener breakdown region, the voltage across the diode is essentially constant at $V_Z$. Therefore, the power dissipated by the Zener diode is $P_Z = V_Z \times I_Z$, where $I_Z$ is the current flowing through the Zener diode. To find the maximum current the Zener diode can safely handle, we use its maximum power dissipation rating and its Zener voltage.

   The maximum current ($I_{Z(max)}$) through a Zener diode is given by:

   $I_{Z(max)} = \frac{P_{max}}{V_Z}$

   where $P_{max}$ is the maximum power dissipation and $V_Z$ is the Zener breakdown voltage.

   Step-by-step Derivation

   1. Identify the given parameters:

      We are given the Zener breakdown voltage ($V_Z$) and the maximum power dissipated ($P_{max}$).

      • Zener breakdown voltage, $V_Z = 7.2 \text{ V}$
      • Maximum power dissipated, $P_{max} = 216 \text{ mW}$

   2. Convert units to SI base units:

      It is good practice to convert all given values to their standard SI units before calculation to avoid errors.

      • $P_{max} = 216 \text{ mW} = 216 \times 10^{-3} \text{ W}$ …

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8. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.For a common emitter transistor amplifier, the voltage across the collector resistance is 2.8 V. If the current amplification factor of the transistor is 70 and the collector resistance is 2 k$\Omega$, then the base current is (A) 20 $\mu$A (B) 30 $\mu$A (C) 10 $\mu$A (D) 40 $\mu$A

   ›Reveal solutionSolution

   The collector current is found from the voltage across the collector resistor, then the base current is obtained using the current gain $\beta = 70$. The base current is 20 μA.

   The key idea here is that in a common emitter amplifier, the collector current $I_C$ is related to the base current $I_B$ by the current amplification factor $\beta$ (also called $h_{fe}$). The voltage across the collector resistor $R_C$ tells us $I_C$ directly via Ohm’s law, and then we simply divide by $\beta$ to get $I_B$.

   1. Find the collector current. The voltage across the collector resistance $R_C = 2\ \text{k}\Omega$ is given as $V_{R_C} = 2.8\ \text{V}$. By Ohm’s law:

   $I_C = \frac{V_{R_C}}{R_C} = \frac{2.8\ \text{V}}{2000\ \Omega} = 0.0014\ \text{A} = 1.4\ \text{mA}.$

   2. Relate collector current to base current. For a common emitter transistor, the current amplification factor $\beta$ is defined as:

   $\beta = \frac{I_C}{I_B}.$

   Here $\beta = 70$, so:

   $I_B = \frac{I_C}{\beta} = \frac{1.4\ \text{mA}}{70} = 0.02\ \text{mA} = 20\ \mu\text{A}.$ …

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9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.The power delivered by the battery shown in the figure is (Take the forward and reverse biased resistances of the diode are zero and infinity respectively) [FIGURE] (A) $2.50~\text{W}$ (B) $5.00~\text{W}$ (C) $7.50~\text{W}$ (D) $3.75~\text{W}$

   ›Reveal solutionSolution

   The two diodes face opposite ways, so only one branch conducts — the one with the $10\ \Omega$ resistor. The reverse-biased branch is an open circuit, so $P = V^2/R = 5^2/10 = \mathbf{2.50~\mathrm{W}}$ — option (A).

   The concept first

   An ideal diode is nothing more mysterious than a switch controlled by the direction of the applied voltage:

   • Forward biased (p-side at higher potential): the depletion layer collapses, resistance $\to 0$ — treat it as a short circuit / piece of wire.
   • Reverse biased (p-side at lower potential): the depletion layer widens, resistance $\to \infty$ — treat it as an open circuit / broken wire. A branch containing a reverse-biased ideal diode can simply be rubbed out of the circuit diagram.

   That is the whole trick of these problems: redraw the circuit with the reverse-biased branches deleted and the forward-biased diodes replaced by wires. What is left is a plain resistor network.

   Here, the two branches hang across the same pair of nodes (they are in parallel) and therefore see the same voltage with the same polarity. But $D_1$ and $D_2$ are drawn pointing in opposite directions. Conventional current can pass through a diode only in the direction of its arrowhead. So for whichever way round the battery is connected, exactly one diode is forward biased and the other is reverse biased — the two can never conduct together. One branch lives; the other is dead.

   Step-by-step

   1. Identify the conducting branch. With the battery polarity as drawn, the branch whose diode arrow points along the direction the battery drives current is forward biased. That is the $D_1$ + $10\ \Omega$ branch. $D_2$, pointing the opposite way, is reverse biased.
   2. Apply the ideal-diode model.

   • $D_1$: forward biased ⇒ $R_{D_1} = 0$ ⇒ replace by a wire. The branch is just $10\ \Omega$. …

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10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The magnetic susceptibility of ferromagnetic materials is (A) $< 0$ (B) $> 1$ (C) $1$ (D) $0$

    ›Reveal solutionSolution

    Ferromagnetic materials have a very large positive magnetic susceptibility (much greater than 1) due to spontaneous alignment of magnetic domains. The correct option is (B).

    Concept & Intuition

    Magnetic susceptibility ($\chi$) measures how easily a material becomes magnetized in an external magnetic field. For ferromagnetic materials (like iron, nickel, cobalt), the internal atomic magnetic moments align strongly and cooperatively, producing a huge magnetization even in a weak field. This is fundamentally different from paramagnetic materials (small positive $\chi$, typically $10^{-5}$ to $10^{-3}$) or diamagnetic materials (small negative $\chi$, about $-10^{-5}$). Ferromagnets have $\chi$ values ranging from $10^2$ to $10^6$ — always positive and much greater than 1.

    Step-by-step reasoning

    1. Recall the definition of magnetic susceptibility

       The magnetic susceptibility $\chi$ is defined by $\mathbf{M} = \chi \mathbf{H}$, where $\mathbf{M}$ is magnetization and $\mathbf{H}$ is the applied magnetic field. For linear materials, $\chi$ is a constant; for ferromagnets, it is nonlinear but always large and positive.

    2. Compare with other magnetic behaviors

       • Diamagnetic materials: $\chi < 0$ (small negative, e.g., copper, water).
       • Paramagnetic materials: $0 < \chi \ll 1$ (small positive, e.g., aluminum, oxygen).
       • Ferromagnetic materials: $\chi \gg 1$ (large positive, e.g., iron has $\chi \approx 200$ at room temperature).

    3. Eliminate incorrect options

       • Option (A) $< 0$: This describes diamagnetism, not ferromagnetism.
       • Option (C) $1$: This would mean the material is exactly as magnetizable as a vacuum (relative permeability $\mu_r = 2$), which is not true for any ferromagnet. …

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11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.The built-in potential of a p-n junction diode is $0.7\,\mathrm{V}$. If the diode is forward biased and the applied voltage is $0.3\,\mathrm{V}$, the effective barrier height is (A) $0.7\,\mathrm{V}$ (B) $0.3\,\mathrm{V}$ (C) $0.4\,\mathrm{V}$ (D) $1\,\mathrm{V}$

    ›Reveal solutionSolution

    The effective barrier height in a forward-biased p-n junction is the built-in potential minus the applied forward voltage, so $0.7\,\mathrm{V} - 0.3\,\mathrm{V} = 0.4\,\mathrm{V}$.

    The key idea is that a p-n junction has a built-in potential barrier (here $0.7\,\mathrm{V}$) that opposes the flow of majority carriers. When we apply a forward bias (positive voltage to p-side, negative to n-side), we are essentially pushing carriers toward the junction, which lowers the barrier. The effective barrier height is therefore the original built-in potential reduced by the applied voltage.

    Why subtract?

    Think of the built-in potential as a hill that carriers must climb. Forward bias is like an external push that helps them up the hill — it effectively reduces the hill’s height. So the net height they face is the original hill minus the push.

    Now, step by step:

    1. Identify the built-in potential

       Given: $V_{\text{bi}} = 0.7\,\mathrm{V}$. This is the barrier with no external voltage.

    2. Identify the applied forward bias

       Given: $V_{\text{applied}} = 0.3\,\mathrm{V}$ (forward bias means the p-side is positive relative to the n-side).

    3. Determine the effective barrier

       In forward bias, the effective barrier height is:

    $V_{\text{eff}} = V_{\text{bi}} - V_{\text{applied}}$

    Substitute:

    $V_{\text{eff}} = 0.7\,\mathrm{V} - 0.3\,\mathrm{V} = 0.4\,\mathrm{V}$ …

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12. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.When an n-type semiconductor is heated (A) number of electrons increases while that of holes decreases (B) number of holes increases while that of electrons decreases (C) number of holes and electrons do not change (D) number of holes and electrons increases equally

    ›Reveal solutionSolution

    In an n-type semiconductor, heating generates equal numbers of electron-hole pairs, so both carrier concentrations increase equally — the correct answer is (D).

    Concept and Intuition

    An n-type semiconductor is doped with donor atoms that provide extra electrons, making electrons the majority carriers and holes the minority carriers at room temperature. When you heat it, you're not just affecting the dopants — you're also providing enough thermal energy to break covalent bonds in the semiconductor lattice itself. Each broken bond creates one free electron and one hole (an electron-hole pair). This intrinsic generation happens equally for both carriers, regardless of the doping type. So while the original majority/minority imbalance remains, both populations rise by the same amount.

    Step-by-Step Reasoning

    1. Understand the initial state

       At room temperature, an n-type semiconductor has many free electrons from donor atoms (e.g., phosphorus in silicon) and relatively few holes. The electron concentration $n \approx N_D$ (donor density), and the hole concentration $p \approx n_i^2 / N_D$, where $n_i$ is the intrinsic carrier concentration.

    2. What happens when temperature increases?

       Heating provides energy that can break covalent bonds in the semiconductor. Each broken bond generates one electron and one hole — this is called intrinsic carrier generation. The number of such pairs increases with temperature.

    3. Effect on electrons and holes

       • Electrons: The original donor electrons are still there, plus the newly generated electrons from bond breaking. So the total electron concentration increases.
       • Holes: The original few holes are still there, plus the newly generated holes from bond breaking. So the hole concentration also increases.

    4. Are the increases equal? …

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