Problem 4.102 Part A Two ideal dc voltage sources electrical conductors that hav

Question

Problem 4.102 Part A Two ideal dc voltage sources electrical conductors that have a resistance of r /m, as shown in the figure. A load having a resistance of R moves between the two voltage sources. Let z equal the distance between the load and the source v, and let L equal the distance are connected by Find z when L = 16 km , ul=1190 V , u2 = 1400 V , R-3.9 , r=5 x 10-5 /m , and u is minimum. Express your answer with the appropriate units. Figure 1 Value Units Submit My Answers Give Up Part B R (movable (+ What is the minimum value of v for the circuit? 01 oad) 02 Express your answer with the appropriate units. Umin-ValueUnits

Explanation / Answer

4.102 : from the given data

A. assuming current throgh R is i, downwards

using kirchoff's loop law

v1 = I(r*x + rx) + iR [ where I is current leaving the v1 voltage source]

in the other loop, similiarly

v2 = I'(r(L-x) + r(L-x)) + iR

also, from junction law

I' + I = i

so I' = i - I

v2 = 2r(i - I)(L - x) + iR

v1 = 2Irx + iR

now given, L = 16,000 m = 16 km

v1 = 1190 V

v2 = 1400 V

R = 3.9 ohm

r = 5*10^-5 ohm/m = 5*10^-2 ohm / km

and v is minimum, where v = iR

so, 1400 = 2*5*10^-2(i - I)(16 - x) + 3.9i

=> 1400 = 0.1(i - I)(16 - x) + 3.9i

1190 = 0.1*Ix + 3.9i

I = (11900 - 39i)/x

hence

1400 = 0.1(i - (11900 - 39i)/x)(16 - x) + 3.9i

1400x = 0.1(ix - (11900 - 39i)(16 - x) + 3.9ix

1400x = 0.1ix - 19040 + 1190x + 62.4i

i = (210x + 19040)/(0.1x + 62.4)

so for i to be minimum

denominator has to be maximum

hence, x = 16

B. when x = 16

i = (210*16 + 19040)/(1.6 + 62.4)

i = 350 A

so v = iR = 1365 V

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