JEE Main 2025 (Online) 3rd April Evening Shift | Semiconductor Question 6 | Physics

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

$$ \text {The truth table corresponding to the circuit given below is: } $$

JEE Main 2025 (Online) 3rd April Evening Shift Physics - Semiconductor Question 1 English

$$ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{~B} & \mathrm{C} \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|} \hline \text { A } & \mathrm{B} & \mathrm{C} \\ \hline 0 & 0 & 1 \\ \hline 1 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|} \hline \text { A } & \mathrm{B} & \mathrm{C} \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{~B} & \mathrm{C} \\ \hline 0 & 0 & 0 \\ \hline 1 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{array} $$

JEE Main 2025 (Online) 3rd April Morning Shift

MCQ (Single Correct Answer)

-1

$$ \text { Choose the correct logic circuit for the given truth table having inputs } A \text { and } B \text {. } $$

Inputs		Output
A	B	Y
0	0	0
0	1	0
1	0	1
1	1	1

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Semiconductor Question 4 English Option 1

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Semiconductor Question 4 English Option 2

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Semiconductor Question 4 English Option 3

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Semiconductor Question 4 English Option 4

JEE Main 2025 (Online) 2nd April Evening Shift

MCQ (Single Correct Answer)

-1

$$ \text { In the digital circuit shown in the figure, for the given inputs the } P \text { and } Q \text { values are : } $$

JEE Main 2025 (Online) 2nd April Evening Shift Physics - Semiconductor Question 7 English

$P=0, Q=1$

$P=1, Q=0$

$\mathrm{P}=0, \mathrm{Q}=0$

$P=1, Q=1$

JEE Main 2025 (Online) 2nd April Morning Shift

MCQ (Single Correct Answer)

-1

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V . For a $400 \Omega$ resistor connected in series, the zener current is found to be 4 times load current. The load current $\left(I_L\right)$ and load resistance $\left(R_L\right)$ are :

$\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$

$\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \Omega$

$\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$

$\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$

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