Semiconductor Electronics

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New answer posted•a year ago

Semiconductor Electronics physics ncert exemplar solutions class 12th chapter two

Two 10cm log, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10^-5N, then separation between the wires is__________cm.

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$F = \frac{μ_{0} \times 2}{4 π} \frac{i_{1} i_{2}}{d} l$

$1 0^{- 5} = \frac{1 0^{- 7} \times 2 \times 5 \times 5}{d} = \frac{10}{100}$

$d = 5 \times 1 0^{- 2} m$

d = 5 cm

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New answer posted•a year ago

Semiconductor Electronics physics ncert exemplar solutions class 12th chapter two

The positive feedback is required by an amplifier to act an oscillator. The feedback here means:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

A portion of the output power is returned back to the input.

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New answer posted•a year ago

Semiconductor Electronics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

14 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

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New question posted•a year ago

Semiconductor Electronics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

<p><strong>The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)</strong></p>

3 Views|Posted a year ago

Asked by Payal Gupta

New answer posted•a year ago

Semiconductor Electronics Class 12th

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100  $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$  and input resistance is 1 $k Ω,$  the value of power gain is x × 10⁴. The value of x is_____________.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} \times \frac{R_{0}}{R_{i}}$

= ${(\frac{10 \times 1 0^{- 3}}{100 \times 1 0^{- 6}})}^{2} \times \frac{2}{1}$

= 2 × 104 = x × 104

= 2

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New answer posted•a year ago

Semiconductor Electronics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 k Ω .$ The voltage gain of the transistor is________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Voltage gain = $\frac{Δ l_{C}}{Δ l_{B}} \times \frac{R_{C}}{R_{B}}$

$\frac{5 \times 1 0^{- 3}}{100 \times 1 0^{- 6}} \times \frac{2}{0.5}$

= 200

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New answer posted•a year ago

Semiconductor Electronics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

The zener diode has a V_z = 30V. The current passing through the diode for the following circuit is…………mA.

15 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$l = \frac{90 ? 30}{4000} = 15 m A$

$I_{1} = \frac{30}{5000} = 6 m A & l_{2} = 9 m A$

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New answer posted•a year ago

Semiconductor Electronics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Take the breakdown voltage of the zener diode used in the given circuit as $6 V$ . For the input voltage shown in figure below, the time variation of the output voltage is: (Graphs drawn are schematic and not to scale)

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Kindly go through the solution

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New answer posted•a year ago

Semiconductor Electronics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

A small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06 T$ experiences a torque of $0.018 N m$ . The minimum work required to rotate it from its stable to unstable equilibrium position is:

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\overset{?}{τ} = \overset{?}{μ} \times \overset{?}{B}$

$\begin{array}{l} \Rightarrow & 0.018 = μ (0.06) (s i n ? 30^{?}) \\ \Rightarrow & μ = 0.6 \\ \Rightarrow & Work = U_{f} - U_{i} \\ \Rightarrow & 2 μ B \\ \Rightarrow & 7.2 \times 10^{- 2} J . \end{array}$