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- For Fixed Bias Configuration Solve for Ib, Ic and le. Atleast 2 decimal place. Be mindful of units. Assume Ic is not equal to le. ... 4 RC 15002 RB VCC 20V 90A/A 100k2 Ib=Blank 1 HA Ic=Blank 2mA le=Blank 3 mA Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answerThe drawback to using this equivalent circuit, however, is that it is defined for a set of operating conditions that might not match the actual operating conditions. a.pi b.hybrid-pi c.hybrid d.re1) The term Bi is defined as Select one: O a. Negative change in reference bus power to the power change DPi on bus i O b. All of these 1/Pfi Clear my choice
- Determine the following for the fixed- biased configuration. (Use 4 decimal places and show complete solution) a. IBQ and ICQ b. VCEQ c. VB , VC and VBCProvide circuit diagram for following equations: AB’{C+D’}’+{G.H}’+{A xor M} {{M+N}’.{Q xnor P’}.{A’.C’}’}Q11. What is LED? Plz Write like a summary
- aloulated m O1 the value you Up = A cos wt +1o Calaulate maarli)-min(i) =? Use constant voltage drop mad. Don7 to write units. . - 3.7V forget14 10 Ω 2 Q 15 2. 1290 16 C D 11 V 1Ω 1 5Ω 10V 13 Ω Dr. Ben Blalock asks that you solve for the currents as shown. If th opposite direction as shown, specify it as a negative current. EQUATION CHECKW1 A 1 IL Tr1 AC Supply V V w2 12 Tr2 use double range voltmeter Fig.(1)
- 23.Compare between 1 ph IM and 3 ph IM. 24.Draw the equivalent circuit of 1 ph IM. 25.Write the equations to determine X0 and RO of 1 ph IM. 26. Why the p.f under no load is a very small value.Determine the following for the fixed- biased configuration. (Use 4 decimal places) a. IBQ and ICQ b. VCEQ c. VB , VC and VBCThe voltage regulation can be equal to zero when: OThe t and current is not too high. OThe p.f. is leading p.f. for any possible current. OThe p.f. is lagging p.f. for rated current.