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Physics-

General

Easy

Question

The part of a transistor which is most heavily doped to produce large number of majority charge carries is :

emitter
base
collector
can be any of the above three

The correct answer is: emitter

Related Questions to study

The energy band gap is maximum in :

The energy band gap is maximum in :

Physics-General

Statement-1 : A transistor amplifier in common emitter configuration has a low input impedence.
Statement-2 : The base to emitter region is forward biased.

Statement-1 : A transistor amplifier in common emitter configuration has a low input impedence.
Statement-2 : The base to emitter region is forward biased.

Physics-General

The current flowing through the zener diode in fig. is-

The current flowing through the zener diode in fig. is-

Physics-General

parallel

Consider the following statements 1 and 2 and identify the correct answer- 1] A zener diode is always connected in reverse bias. 2] The potential barrier of a P-N junction lies between 0.1 to 0.3 V approximately.

Consider the following statements 1 and 2 and identify the correct answer- 1] A zener diode is always connected in reverse bias. 2] The potential barrier of a P-N junction lies between 0.1 to 0.3 V approximately.

Physics-General

Let ABC be a triangle with equations of the sides AB, BC and CA respectively x – 2 = 0, y – 5 = 0 and 5x + 2y –10 = 0. Then the orthocentre of the triangle lies on the line

Let ABC be a triangle with equations of the sides AB, BC and CA respectively x – 2 = 0, y – 5 = 0 and 5x + 2y –10 = 0. Then the orthocentre of the triangle lies on the line

Circles x2 + y2 = 4 and x2 + y2 – 2x – 4y + 3 = 0

Circles x2 + y2 = 4 and x2 + y2 – 2x – 4y + 3 = 0

parallel

The length of intercept on y– axis, by a circle whose diameter is the line joining the points (–4, 3) and (12, –1) is -

The length of intercept on y– axis, by a circle whose diameter is the line joining the points (–4, 3) and (12, –1) is -

The number of common tangents that can be drawn to the circles x²+ y²– 4x – 6y – 3 = 0 and x²+ y²+ 2x + 2y + 1 = 0 is

The number of common tangents that can be drawn to the circles x²+ y²– 4x – 6y – 3 = 0 and x²+ y²+ 2x + 2y + 1 = 0 is

Statement I : $blank to the power of 21 end exponent C subscript 0 end subscript plus blank to the power of 21 end exponent C subscript 1 end subscript plus horizontal ellipsis plus blank to the power of 21 end exponent C subscript 10 end subscript equals 2 to the power of 20 end exponent$ because
Statement II : $blank to the power of 2 n plus 1 end exponent C subscript 0 end subscript plus blank to the power of 2 n plus 1 end exponent C subscript 1 end subscript plus horizontal ellipsis to the power of 2 n plus 1 end exponent C subscript 2 n plus 1 end subscript equals 2 to the power of 2 n plus 1 end exponent$ and ⁿC_r = ⁿC_n_r

Statement I : $blank to the power of 21 end exponent C subscript 0 end subscript plus blank to the power of 21 end exponent C subscript 1 end subscript plus horizontal ellipsis plus blank to the power of 21 end exponent C subscript 10 end subscript equals 2 to the power of 20 end exponent$ because
Statement II : $blank to the power of 2 n plus 1 end exponent C subscript 0 end subscript plus blank to the power of 2 n plus 1 end exponent C subscript 1 end subscript plus horizontal ellipsis to the power of 2 n plus 1 end exponent C subscript 2 n plus 1 end subscript equals 2 to the power of 2 n plus 1 end exponent$ and ⁿC_r = ⁿC_n_r

parallel

If C_r denotes the combinatorial coefficient in the expansion of (1 + x)ⁿ then (nN
Statement 1: $left parenthesis n plus 1 right parenthesis open parentheses C subscript 0 end subscript superscript 2 end superscript plus C subscript 1 end subscript superscript 2 end superscript plus C subscript 2 end subscript superscript 2 end superscript plus horizontal ellipsis plus C subscript n end subscript superscript 2 end superscript close parentheses greater than open parentheses C subscript 0 end subscript plus C subscript 1 end subscript plus C subscript 2 end subscript plus horizontal ellipsis plus C subscript n end subscript close parentheses to the power of 2 end exponent for all n greater or equal than 2 comma n element of N$ because
Statement 2: Root mean square of n distinct real numbers is always greater than their arithmetic mean, n ≥ 2

If C_r denotes the combinatorial coefficient in the expansion of (1 + x)ⁿ then (nN
Statement 1: $left parenthesis n plus 1 right parenthesis open parentheses C subscript 0 end subscript superscript 2 end superscript plus C subscript 1 end subscript superscript 2 end superscript plus C subscript 2 end subscript superscript 2 end superscript plus horizontal ellipsis plus C subscript n end subscript superscript 2 end superscript close parentheses greater than open parentheses C subscript 0 end subscript plus C subscript 1 end subscript plus C subscript 2 end subscript plus horizontal ellipsis plus C subscript n end subscript close parentheses to the power of 2 end exponent for all n greater or equal than 2 comma n element of N$ because
Statement 2: Root mean square of n distinct real numbers is always greater than their arithmetic mean, n ≥ 2

The sum $stretchy sum from k equals 1 to n minus gamma plus 1 of k to the power of n minus k end exponent C subscript gamma minus 1 end subscript$ is equal to

The sum $stretchy sum from k equals 1 to n minus gamma plus 1 of k to the power of n minus k end exponent C subscript gamma minus 1 end subscript$ is equal to

$stretchy sum from i equals 0 to n of stretchy sum from j equals 0 to n of stretchy sum from k equals 0 to n of open parentheses fraction numerator n over denominator i end fraction close parentheses open parentheses fraction numerator n over denominator k end fraction close parentheses comma open parentheses fraction numerator n over denominator r end fraction close parentheses equals blank to the power of n end exponent C subscript r end subscript$

$stretchy sum from i equals 0 to n of stretchy sum from j equals 0 to n of stretchy sum from k equals 0 to n of open parentheses fraction numerator n over denominator i end fraction close parentheses open parentheses fraction numerator n over denominator k end fraction close parentheses comma open parentheses fraction numerator n over denominator r end fraction close parentheses equals blank to the power of n end exponent C subscript r end subscript$

parallel

The expression $open parentheses table row cell n end cell row cell k end cell end table close parentheses plus open parentheses table row cell n end cell row cell k plus 1 end cell end table close parentheses$ where $n greater or equal than k greater or equal than 1$ is the same as

The expression $open parentheses table row cell n end cell row cell k end cell end table close parentheses plus open parentheses table row cell n end cell row cell k plus 1 end cell end table close parentheses$ where $n greater or equal than k greater or equal than 1$ is the same as

Which of the following pairs of ions cannot beseparated by H₂Sindilute HCl?

Which of the following pairs of ions cannot beseparated by H₂Sindilute HCl?

Chemistry-General

The pairofcom pounds which cannot exist together in solution is:

The pairofcom pounds which cannot exist together in solution is:

Chemistry-General

parallel

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