Question

Problem 4. A 10 mH inductor has a sudden current change from 200 mA to 100 mA in 1 ms. Find the induced voltage. Problem 5. A induced voltage across a 10 mH inductor is v(t) 120 cos (377t) V. Find (a) the expression for the inductor current and (b) the expression for the power. The current in a 25 mH inductor is given by the expressions: i(t) 0 i(t) 10 (1-e) mA Find the voltage across the inductor and the energy stored in it after 1 s. Problem 6. t<0 t>0 Problem 7 current source. Find (a) the equivalent inductance and (b) the total energy stored. A 20 mH inductor and a 12 mH inductor are connected in series with a 1 A Problem 8. Determine the total capacitance Cr of the network shown below. 1 wious MacBook Air 00 F3 F4 FS F7 F10

Answer 1

Induced voltage = v= L L=10mH 1 - 200 mA 12 = 100mA At = 1 msec = 10 xud [100 - 2000 2003 J 1x10 1 -100 o molts -1 volt L=10mH Y(t) = laocos (377) a) Inductor custent = ilt): Svet) dt Ein S126.cos (32+ k) dt 16 X10 = FE) = 12x103 sin (377) 31-83 sin (347t) amps 377 5) Expression for power = f(t) = v(t) ilt) [20 cos(377 +) ) [3193 sin (2+1)] = P(t) = 3819.6 [a cos (377t) sin(3114) le(t) = 1909-8 sin (754)t VA) 2 L=25mt ; ilt)=0 tao (42=10[1-2 t]mat so VL) - L.di(l) - 25x103 d 101-et] dt he] at

V(t) = 0.25 1o- e Juli) = 0.250 tmu too lure) = 0 co b) energy stored in it after 1 sec is energy stored at 1 sec ()10[1-6'] = 6.32 mA. energy LIP = 4x 25 x 109 x (6.32 x 10-3)? = 499.28 x109 ] = 0.499x10-6 J Enagy = 0.499 MJ | = 0.499 M J L = 20mH ; When inductors are connected in series L2=12mh .. a) then equivalent inductance = Leq = 1,tha Equivalent inductance = ( 20 +12) mH Theq = 32 ml] b) Total Energy stored = 1/2 LI I 1:1= 14) = ** 32x10-3x (1)? - TomJ
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