Solution of ECE 300 Test 1 F09
1. Let Vs = 10V, R = 20, I = 2 A . Find the numerical power delivered to R and Z. (The voltage across each of these three elements connected in parallel is the same.)
PR = V 2 / R = (10V) / 20 = 5W , Pz = 10V ( 2 A )
Solution of ECE 300 Test 6 F09
1. Find the numerical energy stored in the capacitor and in the inductor.
V1 = 24V , R2 = 18 , C3 = 1F , L4 = 100mH
The source voltage is constant. Therefore all currents and voltages are constant. Therefore t
Solution of ECE 300 Test 7 F09
1. For the circuit below, fill in the blanks below with numbers.
I1 = 3A , R3 = 5 , R4 = 15 , L5 = 100 mH
() i (0 ) = 0 A i (0 ) = 3 A v (0 ) = 0 V
is 0 = 0 A
4 L L
() i ( 0 ) = 3 / 4 A i (0 ) =
Solution of ECE 300 Test 8 F09
1. In the space provided graph this voltage versus time. Put a scale on the vertical axis so that actual voltages could be read from the graph. v (t ) = 4 u ( t ) 6 u ( 2 t ) + 5 u (t 4 )
2. Find the num
Solution of ECE 300 Test 9 F09
1. With reference to the circuit below, find numerical values for the following.
vC 0 + = _ V , i L 0 + = _ A vL 0 + = _ V , iC 0 + = _ A
d d ( iL (t ) + = _ A/s , dt ( vC (t ) + = _ V/s dt t =0 t =0 d d ( iC (t ) + = _ A/s
Solution of ECE 300 Test 11 F09
1. With reference to the circuit below, the phasor voltages of the voltage source, resistor and inductor are represented in the phasor diagram. The circuit is operating in the sinusoidal steady state with = 120 . If the imp
Solution of ECE 300 Test 12 F09
1. An industrial load operating at 60 Hz can be modeled as a resistance of 20 in series with an inductance of 300 mH.
The impedance is Z = R + j L = 20 + j ( 377 / s ) (0.3 H ) = 20 + j113.1 = 114.85 79.97 . ratio of voltag
If 60 C of charge pass through an electric conductor in 30 seconds, determine the current in the conductor.
i= Q 60 = =2A t 30
If the current in an electric conductor is 2.4 A, how many coulombs of charge p
Solution of ECE 300 Test 3 F09
1. In the circuit below R1 = 5 , R2 = 15 , R3 = 20 , R4 = 10 , Vs = 20V and I s = 1A . (a) The current I x can be expressed in terms of the node voltages v1 , v2 and v3 . Fill in each blank in the equation with a single numb
Solution of ECE 300 Test 4 F09
1. Draw an equivalent circuit at the terminals a and b consisting of one independent voltage source in series with one resistor. Be sure to indicate the numerical voltage of the source and the numerical resistance of the res
Solution of ECE 300 Test 5 F09
1. In the circuit below, find the numerical value of Vout . V1 = 3V , V2 = 5V , R1 = 10 k R2 = 5 k , R f = 20 k
R2 V2 V
Rf Rf V V Vout = V1 + V2 = R f 1 + 2 R1 R2 R1 R2
5 20 3 6 Vout = 20, 000 + = 20, 000 + =