Class 12th

Contributor-Level 10

$A s V^{2} = V_{R}^{2} + C_{C}^{2}$

$? V_{R} = \sqrt[]{V^{2} - V_{C}^{2}} = \sqrt[]{10^{2} - 8^{2}} = 6 V$

$\tan ? θ = \frac{X_{C}}{X_{R}} = \frac{V_{C}}{V_{R}} = \frac{8}{6} = \frac{4}{3}$

Now,

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In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt. The collector current changes by 0.05 mA The output resistance will be

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Class 12th Physics Wave Optics

Contributor-Level 9

Here $Δ V_{c} = 0.5 V, Δ i_{c} = 0.05 m A = 0.05 * 1 0^{- 3} A$

Output resistance is given by

$R_{o u t} = \frac{Δ V_{c}}{Δ i_{c}} = \frac{0.5}{0.5 * 1 0^{- 3}} = 1 0^{4} Ω = 10 k Ω .$

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a year ago

For the setup shown in figure, the distance between slits much smaller as compared to the distance of slits from the screen. The total number of minimas on the screen are (where $d = 4 λ)$

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Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

For minima

$d \cos ? θ = (n + \frac{1}{2}) λ$

$? \cos ? θ = \frac{2 n + 1}{8}$

$? n = - 4, - 3, - 2, - 1, 0, 1, 2, 3$

$? 8$

minimas

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a year ago

Two long straight conductors carry currents $3 A$ and $1 A$ into the plane of paper. A circular path is imagined to enclose these currents. The value of $\oint \vec{B} . \vec{d l}$ is

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Contributor-Level 10

From ampere's circuital law

$? \overset{?}{B} . \overset{?}{d l} = μ_{0} i_{e n c}$

$= μ_{0} (3 + 1) = 4 μ_{0}$

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a year ago

In a transistor if $\frac{I_{C}}{E_{E}} = α$ and $\frac{I_{C}}{I_{B}} = β$ . If $α$ varies between $\frac{20}{21}$ and $\frac{100}{101}$ , then the value of $β$ lies between

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Contributor-Level 10

The relation between $?$ and $?$ is given as

$? = \frac{?}{1 - ?}$

$?_{1} = \frac{\frac{20}{21}}{1 - \frac{20}{21}} = 20$

$?_{2} = \frac{\frac{100}{101}}{1 - \frac{100}{101}} = 100$

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a year ago

For the transistor circuit shown below, if $β = 100$ , voltage drop between emine r and base is 0.7 V then value of $V_{C l}$ will be

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Contributor-Level 9

$i_{b} = \frac{5 - 0.7}{8.6} = 0.5 m A$

$\Rightarrow I_{c} = β I_{b} = 100 * 0.5 m A$

By using

$V_{C E} = V_{C C} - I_{C} R_{L} = 18 - 50 * 1 0^{- 3} * 100 = 13 V$

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a year ago

The charge on a hole is equal to the charge of

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Contributor-Level 9

The charge on hole is positive.

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a year ago

An electron of mass ' $m$ ' with an initial velocity $\vec{v} = v_{0} \hat{i}$ is in electric field $\vec{E} = E_{0} \hat{j}$ . If $λ_{0} = \frac{h}{m v_{0}}$ , its de Broglie wavelength at time ' $t$ ' is given by (Check the symbols for vectors)

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Contributor-Level 10

$\overset{?}{F} = e E_{0} \overset{?}{j}$

${\overset{?}{a}}_{y} = \frac{e E_{0}}{m} \overset{?}{j}$

$= v_{0} \overset{?}{i} + (\frac{e E_{0} t}{m}) \overset{?}{j}$

$|\overset{?}{v}| = \sqrt[]{v_{0}^{2} + {(\frac{e E_{0} t}{m})}^{2}} = v_{0} \sqrt[]{1 + {(\frac{e E_{0} t}{m})}^{2}}$

$λ = \frac{λ_{0}}{\sqrt[]{1 + {(\frac{e E_{0} t}{m})}^{2}}}$

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a year ago

The nucleus 115/48cd, after two successive β?decay will give

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Contributor-Level 9

Two successive β decays increase the charge no. by 2.

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a year ago