Electronics

Step 1: Understanding the Concept:
In a Bipolar Junction Transistor (BJT), leakage current is the small current that flows even when the transistor is supposed to be "off". This current is temperature-dependent. The leakage current is defined differently for common emitter (CE) and common base (CB) configurations.
: Collector-to-Emitter leakage current with the base open (for CE configuration).
: Collector-to-Base leakage current with the emitter open (for CB configuration).
These two leakage currents are related through the transistor's current gain parameters.
Step 2: Key Formula or Approach:
The relationship between the common emitter leakage current ( ) and the common base leakage current ( ) is given by:
where is the common emitter DC current gain. We are given and and need to find .
Step 3: Detailed Explanation:
1. Identify the given values:
Common emitter leakage current, .
Common emitter current gain, .
2. Rearrange the formula to solve for :
3. Substitute the values and calculate:
Step 4: Final Answer:
The calculated value for the common base leakage current is approximately 2.29 A. The closest option is 2.28 A.

Step 1: Understanding the Concept:
Peak Inverse Voltage (PIV) is the maximum reverse voltage that a diode in a rectifier circuit must be able to withstand without breaking down. For a simple half-wave rectifier, during the negative half-cycle of the input AC voltage, the diode is reverse-biased. The maximum voltage it experiences is the peak voltage of the AC waveform at the secondary of the transformer.
Step 2: Key Formula or Approach:
1. Calculate the RMS voltage at the secondary of the transformer ( ). 2. Calculate the peak voltage at the secondary ( ). 3. For a half-wave rectifier, PIV = .
The relevant formulas are:
Step 3: Detailed Explanation:
1. Find the secondary RMS voltage ( ):
The primary voltage is given as V. The turns ratio is given as 20:1, which means . 2. Find the secondary peak voltage ( ):
The relationship between peak voltage and RMS voltage for a sinusoidal AC supply is:
3. Determine the Peak Inverse Voltage (PIV):
In a half-wave rectifier, during the negative cycle, the diode is open and the full secondary peak voltage appears across it.
Step 4: Final Answer:
The peak inverse voltage for the diode in this circuit is 15.554 V. This corresponds to option (D).

Step 1: Understanding the Concept:
This question tests the knowledge of the primary application or characteristic property of various semiconductor devices.
Step 2: Detailed Explanation:
Let's analyze each device in List-I and find its corresponding match in List-II.

(A) Zener diode: A Zener diode is specifically designed to operate in the reverse breakdown region. In this region, it maintains a nearly constant voltage across its terminals regardless of the current flowing through it. This property makes it ideal for use as a voltage regulator. So, (A) matches with (II).

(B) Tunnel diode: A tunnel diode is a heavily doped p-n junction diode that exhibits the quantum mechanical effect of tunneling. Its most notable characteristic is a portion of its forward-bias I-V curve where the current decreases as the voltage increases. This is known as a negative resistance region. This property is utilized in making very high-frequency oscillators. So, (B) matches with (I).

(C) Rectifier: A rectifier is an electronic circuit that converts alternating current (AC) to direct current (DC). The output of a simple rectifier is not a steady DC but a pulsating d.c. voltage, which contains a DC component and some AC ripple. A filter is then used to smooth this output. So, (C) matches with (III).

(D) Light emitting diode (LED): An LED is a semiconductor device that emits light when an electric current passes through it. The color of the light depends on the semiconductor material used. Gallium Arsenide Phosphide (GaAsP) is a common material used to make red, orange, and yellow LEDs. So, (D) matches with (IV).
Step 3: Final Answer:
The correct pairings are:
(A) - (II)
(B) - (I)
(C) - (III)
(D) - (IV)
This corresponds to option (A).

Concept: In a series LCR circuit, the phase difference between current and applied emf depends on resistance and reactance.
Step 1: Understanding reactance.
Inductive reactance:
Capacitive reactance:
Step 2: Net reactance.

Step 3: Phase angle formula.

Step 4: Evaluating the options.

• Correct
• Incorrect
• Incorrect
• Incorrect

Step 5: Conclusion.
Thus, the phase angle is given by .

Concept: Efficiency of a Carnot engine depends only on the temperatures of the hot and cold reservoirs.
Step 1: Formula for Carnot efficiency.

Step 2: Substituting values.

Step 3: Calculation.

Step 4: Convert to percentage.

Step 5: Evaluating the options.

• 26.8% Correct
• 36.8% Incorrect
• 46.8% Incorrect
• 56.8% Incorrect

Step 6: Conclusion.
Thus, the efficiency of the Carnot engine is 26.8%.

Concept: Maxwell's equations describe the behavior of electric and magnetic fields. Gauss's law for magnetism states that there are no magnetic monopoles.
Step 1: Understanding Gauss's Law in Magnetostatics.
It states that the net magnetic flux through a closed surface is zero.
Step 2: Mathematical expression.
This is represented as:
Step 3: Physical meaning.
Magnetic field lines always form closed loops; they do not begin or end like electric field lines.
Step 4: Evaluating the options.

• Gauss's law for electricity (incorrect)
• Gauss's law for magnetism (correct)
• Faraday's law (incorrect)
• Ampere's law (incorrect)

Step 5: Conclusion.
Thus, Gauss's Law in magnetostatics is .