Physics-
General
Easy

Question

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

  1. 20 mA    
  2. 25 mA    
  3. 15 mA    
  4. 5 mA    

The correct answer is: 5 mA

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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
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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

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

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

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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

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Maths-General
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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 exponentbecause
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 exponentand nCr = nCn 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 exponentbecause
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 exponentand nCr = nCn r

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If Cr 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 Nbecause
Statement 2: Root mean square of n distinct real numbers is always greater than their arithmetic mean, n ≥ 2

If Cr 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 Nbecause
Statement 2: Root mean square of n distinct real numbers is always greater than their arithmetic mean, n ≥ 2

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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 subscriptis equal to

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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

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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 parentheseswhere n greater or equal than k greater or equal than 1is the same as

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Maths-General
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Chemistry-

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

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Thpairofcom pounds which cannot exist together in solution is:

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If the greatest value of the term independent of x in expansion of open parentheses x s i n invisible function application p plus x to the power of negative 1 end exponent c o s invisible function application p close parentheses to the power of 10 end exponentis achieved at P equals thetaThen the locus of point from which pair of tangents be drawn to x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4including an angle thetais

If the greatest value of the term independent of x in expansion of open parentheses x s i n invisible function application p plus x to the power of negative 1 end exponent c o s invisible function application p close parentheses to the power of 10 end exponentis achieved at P equals thetaThen the locus of point from which pair of tangents be drawn to x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4including an angle thetais

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The sum of the series fraction numerator 5 over denominator 1.4 to the power of 2 end exponent end fraction plus fraction numerator 11 over denominator 4 to the power of 2 end exponent times 7 to the power of 2 end exponent end fraction plus fraction numerator 17 over denominator 7 to the power of 2 end exponent times 10 to the power of 2 end exponent end fraction plus horizontal ellipsis.. plusis equal to

The sum of the series fraction numerator 5 over denominator 1.4 to the power of 2 end exponent end fraction plus fraction numerator 11 over denominator 4 to the power of 2 end exponent times 7 to the power of 2 end exponent end fraction plus fraction numerator 17 over denominator 7 to the power of 2 end exponent times 10 to the power of 2 end exponent end fraction plus horizontal ellipsis.. plusis equal to

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In the expansion of open parentheses x plus x to the power of 2 end exponent plus horizontal ellipsis horizontal ellipsis.. close parentheses open parentheses 1 plus x plus x to the power of 2 end exponent plus x to the power of 3 end exponent close parentheses open parentheses x to the power of 2 end exponent plus horizontal ellipsis horizontal ellipsis. plus x to the power of 10 end exponent close parenthesesthe coefficient of is

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