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3 years ago

How is p-n-p transistor biased to operate in the active mode

Physics

1 answer:

STALIN [3.7K]3 years ago

5 0

Answer:

Shown from explanation.henc2 we

Explanation:

For the transistor to work in the active mode meaning for it to conduct normally, the p- side has to the connected to the positive side of the battery and the n- side the negative terminal of the battery.

Hence we have two batteries such that both ends are connected to the positive terminal and the negative side connected to the middle the n- side of the transistor.

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andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

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There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

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Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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Have you tried to look this up on Quizlet. I just had a similar question on one on my quizzes & the whole test was on Quizlet

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Solnce55 [7]

Answer:

Explanation:

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Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.

The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.

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Dimas [21]

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How is p-n-p transistor biased to operate in the active mode​

How is p-n-p transistor biased to operate in the active mode