Wr |db 25 N R2 I ab 15 N R1 Icb 15 V 18 V + 12 V 20 Ω R4 10 N R5 Ibd 30 N R3 I ba I bc I ac I dc I cd d I ca a C 30 N R6 50 Ω R, I ad I da 14 V 5Ω: Rg + 15 N R10 8 V 8Ω Ro
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Compute for the following current branch using KCL and KVL
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- Capacitive Circuits Fill in all the missing values. Refer to the formulas that follow. XC=12fCC=12fXCf=12CXc Capacitance XC Frequency 38 F 60 Hz 78.8 400 Hz 250 pF 4.5 k 234 F 10 kHZ 240 50 Hz 10 F 36.8 560 nF 2 MHz 15 k 60 Hz 75 nF 560 470 pF 200 kHz 6.8 k 400 Hz 34 F 450Determine the following DC: Vg Vgs Idq Ie Ib Ic Vd Vc Vb Vds In AC: Avs Vo Zin ZoTable 9.1 Calculated Measured Is(p-p) 15.05mA∠-57.86° 15mA IL(p-p) 12.732mA∠-90° 12.6mA IR(p-p) 8mA∠0° 8mA VRs(p-p) (for Is) 150mV 150mV VRs(p-p) (for IL) 126mV 126mV
- Determine:(a) Zin(b) Zo(c) Io if Vs=0.5mVp-p(d) Avs=Vo/VsAn impedance of 100 - j50 ohms in a network operating at 50 Hz represents a) A resistor of 100 ohms and an inductor of 0.159 Henrys b) A resistor of 100 ohms and a capacitor of 63.66 microfarads c) A 100 ohm resistor and a 1 Henry inductor d) A 100 ohm resistor and a 400 microfarad capacitor e) None of the aboveMeterology: How do you mathematically derive this modified refractivity equation? M_rf = (77.6/T)*(P + 4810e/T) + 0.157z
- Solve for B to E: Answers: a. 2247.8405 Ab. 292.0629 kVc. 2291.3829 Ad. 667.10875 MW, 53.2164 MVAR (cap), 669.228 MVAe. 32.89125 MWAvresistor of 51.0 Q, an inductor of 22.0 pH and a capacitor of 150 pF are in parallel. The frequency is 1.00 MHz. Whatis the complex impedance, R_jX? A.51.0_j14.9. B.51.0_/149 C.46.2_j14.9. D.46.2_j14.9. ¥ Show Answer Answer: DSIGNAL POWER of v(t)( NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE).