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Determine the furrent drawn from a 12 V suly with internal resistacne 0.5Omegaby the infinite network shown in Each rsistor has 1Omega resistance. |
Answer» SOLUTION :  Here no. of identical UNITS on the left side of line CD and line PQ are same and so equivalent resistance of given infinite LADDER between points P and Q as well as between points C and D must be same . Let it be x . thus , `R_(PQ) = R_(CD) ` = x ... (1) Now, let us remove part on the left side of line CD and let us replace it by equivalent resistance x.  Now, equivalent resistance between points P and Q in above figure is, `R_(PQ) = 1 + ((x xx 1)/(x + 1) ) + 1 ` `therefore x =2 + (x)/(x +1) `[ from equation (1)] `thereforex (x + 1) = 2 x (x + 1) + x ` `thereforex^(2) + x = 2x + 2 + x ` ` therefore x^(2) - 2x - 2 = 0 ""` .... (2) Comparing above equation with`ax^(2) + bx + c = 0 ` we get a = 1, b = - 2, c = -2 Now,`Delta = b^(2)- 4ac ` = `(-2)^(2) - 4 (1) (-2)` = 4 + 8 `Delta = 12 = 4 xx 3"" rArr sqrt(Delta ) = 2 sqrt(3)` Solution of equation (2) will be, ` x = (- b pm sqrt(Delta ))/( 2 a ) = - ((-2) pm 2 sqrt(3))/(2(1))` ` therefore x = (2 + 2 sqrt(3))/(2 )"" (because x gt 0 )` `therefore x =(2(1 + sqrt(3)))/(2 ) = 1"" sqrt(3) = 1 + 1.732 ` `therefore ` x =2.732 `Omega ""` .... (3) Now, as per the statement above resistance is connected across a battery (12 V, 0.5 `Omega` ) and so , For adjoining loop, `I = (epsilon)/(x +r)` `therefore I = (12)/(2.732 + 0.5) ` ` therefore I = (12)/( 3.232)` `therefore I = 3.713 ` A 
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