circuits, pleaseeeee solve questionnn 19 2A (4)le4.8amQ12) Neper frequency a isa) 4 rad/s b) 16 rad/s c) 25 rad/s d) 2 rad/s e) 5 rad/sQ13) Resonant frequency wo isa) 4 rad/s b) 5 rad/sc) 4 rad/sd) 16 rad/s e) 25 rad/sQ14) The voltage v (t=1s) can be found asa) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V e) 12.61 Va) R=0.22, C=0.2 Fd) R=192, C=1Fm1.92 (234149Fig. 6Fig.7Q10) The average power supplied by the dependent source of Fig. 6 can be determined asa) 5Wb) 24 Wc) 96 Wd) 192 Wc) 48WQ11) The Va for the the circuit shown in Fig. 7 as seen from the terminal a-b can be found as:a)-/220 Vd)-110 V )-165 Vb)-/330 Vc)-j55 VPlease refer to the circuit of Fig. 8 for the following questions (Q12, Q13 and Q14). Assumingit(0)-8 A, and V.(0)-40 V.tethLmLT1.61, 803034Lm1520 AZFig. 9Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is:a) 1/2 Ld) 5/8 L e) 4/7 Lb) 4/9 Lc) 7/4 LQ19) For the circuit shown in Fig. 11, the value of C needed tomake the response underdamped with unity damping factor (a= 1) is:SH6 MA24 V +L3k0zwww492-NN202ΣΕΩΣ1022t=03kQFig. 82mFHF210ww330455+Fig. 10SkΩ Σ4 m²Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is:a) 32 mJd) 8 mJ e) 16 mJb) 128 mJc) 256 mJQ17) If v(t)=15 cos(1000t+66°) V and i(t)-2cos(1000t+450°) A, then v(t) leads i(t) bya) 156°b) 0-24°c) 204°d) 24°e) 66°Q18) Assuming that the input impedance is given as Zin= 1+j1 22 and o-1 rad/s, then the inputadmittance can be represented as the parallel combination of:b) R=0.2 02, L=0.2 He) R=202, L=2Hc) R=10, L=1H100 0.5H C= 10 mFFig. 12a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mFThe switch in Fig. 12 has been in position A for long time. At t=0, the switch moves to position B.Please refer to the circuit of Fig. 12 for the following questions (Q20, Q21 and Q22)Fig. 11Q20) v(0) can be found as:d) 15 Ve) 24 Va)30 V b) 12.5 V c) 9VQ21) v(co) can be found as:a) 24 V b)30 V c) 12.5 V d) 9VQ22) The voltage v(t) at t = 1 s is:b)24.9 V c) 30 Ve) 15 Va) 20.9 Vd) 39.1 Ve) 27.97 VB04f12.5m.9.51,4kQwwwP0.5 mFb41030 V The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit ofFig. 1 for the following questions (Q1, and Q2)Q1) The time constant t can be found as:a) 6.67 s b) 0.3 sc) 10 sQ2) The current i(t) at t- Im s is:a) 2.02 A b) 6 Ac) 4.02 Aa) 1.23 cos(10t +30°) Vd) 2.25 cos(10t-53.6%) Vd) 0.1 s e) 0.15 sFig. 1Refer to the circuit of Fig. 2 for the following 3 questions (Q3, Q4 and Q5)Q3) By using superposition technique, the contribution of the 2cos10t voltage source to the value ofvi(t) is:a) 12.5cos(500t - 0.107°)d) 12.5 cos(500t + 89.9°)d) 5.98 A e) 4 AQ5) The value of the inductance of the j2 02 impedance is:a) 0.2 H b) 10 Hc) 20 H d)1.6 He) 16 H2cos101 VQ7) The current in(t) of Fig. 4 can be found as (mA):2002 -4.0wwwHEb) 1.23 cos(10t-30%) Ve) 1.79 cos(10t-26.57°) VQ4) By applying KCL to the node v/(t), the value of the voltage labeled vi(t) is (V):a) 2.86 cos(10t +77.9°)b) 2.86 cos(10t-77.9°)c) 4.1 cos(10t +62.3°)d) 4.1 cos(10t-62.3°)e) 3.92 cos(10t +77.9°)f) 3.92 cos(10t -77.9%)Q6) Referring to the circuit of Fig. 3, Zen can be determined as:a) 22+j6 2 b) 18+j62 c) 22-j62 d) 18-j62 e) -18+j6paa)-823.5-j294.1d)-751.3+j457.75525 cos 10rVb) 12.5cos(500t+0.107°)e) 12.5 cos(500t + 0.205°)10.0220:2c) 2.25 cos(10t +53.6%) V1.79 cos(10t+26.57°) V15021=0Q9) The complex power absorbed by voltage source is (VA)b)-751.3-j457. c)-823.5+j294.1e) 751.3-j457.7Fig. 2Fig. 3Refer to the circuit of Fig. 5 for the following 2 questions (Q8, and Q9)Q8) The current through the-j10 2 can be found as (Arms)a) 8.75/19.65*b) 8.752-19.65*c) 10.25290*d) 10.25Z-90°e) 202-53.26*f) 20253.261.5 Hmtacos5001 Vc) 12.5cos(500t - 89.9°)f) 12.5 cos(500t - 0.205°)100/2100/0 V ms10520.21Fig. 446Fig. 50.3 mH720 (2m2002www10

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Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The cuida ni 10 Q www 0.5 H ell C= 10 mF