9:45 PM o 33% 1 of 2 EECE 200/201 Spring 2017 -ASSIGNMENT 1 DUE: Monday, January

Question

9:45 PM o 33% 1 of 2 EECE 200/201 Spring 2017 -ASSIGNMENT 1 DUE: Monday, January 30, 2017 Please hand in the assignment at the beginning of the session, hand written and not typed. P1.1 The voltage drop v V across a certain device, and the current i A through it in the direction of the voltage drop, are related by: i-8-22. vs 0 and 2 V i-0. (a) Determine the power absorbed by the load when v 1V, and when v 2 V: (b) at what value of vis the instantaneous power a maximum? (c) If v(t)-2e V, t20 s, what is the total charge that passes through the device from t 0 to t 2 s? P1.2 Determine in Figure P2.2.6 the voltage across each current source, the current 0.8V through each voltage source, and the 10A power delivered or absorbed by 20 V each source. 0.5 Figure P22.6 P1.3 Determine the power delivered or absorbed 00 mA by the dependent source in Figure P2.323. 0 V 2V, 0 V 100 5 V Figure P2.3.23 100

Explanation / Answer

Q1.1

power=i*v

when v=1 volts, i=8-2*v^2=6 A

so power=6*1 W=6 W

when v=2 volts

i=8-2*2^2=0 A

power =0 W

part b:

instantaneous power is non-zero for 0<v<2 only

instantenous power in this range=P=i*v=(8-2*v^2)*v=8*v-2*v^3

taking derivative w.r.t. v and equating to zero,

we get

dP/dv=0

==>8-6*v^2=0

as d^2P/dv^2=-12*v is always negative for v>0, solution for dP/dv will give the maxima .

==>v=sqrt(8/6)=1.1547 volts

then maximum value of instantaneous power=8*1.1547-2*1.1547^3=6.1584 W

part c:

v(t)=2*e^(-t) volts, for t>=0

then current=8-2*v^2=8-2*4*e^(-2*t)=8-8*e^(-2*t)

charge =integration of i*dt

=integration of (8-8*e^(-2*t))*dt

=8*t+4*e^(-2*t)

using limit from t=0 to t=2 seconds,

total charge=8*2+4*e^(-2*2)-(8*0+4*e^(-2*0))=12.0732 C

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