Question

d. none of the above The following questions pertain to the circuits shown below. 3mA ? 30 MO 15 M2 бmA In an amplifier design, your company has a non-ideal current source. The amplifier acts like a resister of 15MQ but can only take 3mA as input or it will overheat and cease function. It was discovered in testing that the above schematic delivered 4mA to the amplifier and most of the amps burned out. Over 1000 of these amps have already been built with such tight space requirements that there is only enough room for one component to be placed in parallel with the amplifier Final Exam, Spring 2019 ENEE 3518 (5pts) 17. The company is also considering replacing the non-ideal current source with a cheaper one that delivers 5mA and has an internal resistance of 15M. Will this current source work in this design? Yes/No (Circle one) (5pts) 18. What component needs to be placed in the location of the question mark? a. A 10MQ resistor 6. Α 15ΜΩ resistor c. A 30M2 resistor d. A 60M2 resistor for all possible values of A

Answer 1

4 mA Gineust 12mA. GMAT Boma 15MA Page No Date: / dous 17) for the ginen question the circuit diagram : as followed ylz now from current distribution, 5mA We wane 15mR 152 2.5 2.5mA.. D2 &mA x 45 3antar Z ! I = I 2 = 2.5mA. As it is genen that only 2 mA is acceptable so the answer for question in 17 is No.

method. Page No. passing b loXTS. tous Step 31- vuurres for 18) we need to go through case study nie we need to check each and every option & then career find out in which 3 mA current through the 15m or resistance (a) lonry 2 Steppt equinalent lollis = bhur Steps - Current through 6 Mr = 628 25 mA Steps - noltage across 6 M2 = 6XE = 30 Step 4: - current through mar 30% 2 mA > No (6) 15 m2 Step 1: - equivalent: 15/115 Step2; - current through tomar - 48 mA Step 2 - 0 Moltage across 7.5MR = 4.8*7.5 = 36 Step 4 : - current through 15 msa 36/15 -2.0mA NO (6) 30m2 Step 1 - equinalent a - 4.5mA Step 27- current throyli lomba across loma 4.5000 45 u. step 4 - aleaga through 45/1523m. A sYes. Hence option (C) is the tonight option. TSXIS 25.tis 7.5 MR. 3ollis loma 6x30 40 15m a Also, atb - In step we find equinalent resistance of two parallel resistan a&b as 7 Du Step tind current through current díuision will.. ie in 1x Yath → Qu Step 3, we find noltage by D&R In Step 4, we find current as we know for parallel connection no noltage are same, and then Da VIR

Q17.) ANSWER - NO

Q18.) ANSWER - OPTION (C) = 30M$\Omega$