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An operational amplifier connected in a noninverting configuration hasan open−loop gain of
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Chapter 14 Solutions
MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
- I need help with this problem and an explanation of the solution for the image described below. (Introduction to Signals and Systems)arrow_forwardDesign a battery pack for an electric bike that consumes in average 10Wh/mile and drive 30 miles per charge. The battery state of charge window is 80%. Design the battery by using new commercial cylindrical cells with 20mm diameter and 80mm height. The battery is constructed based on graphite anode C6 and cathode Li(Ni0.8Co0.15Al0,05)O2 that provides 3.75V at the cell level and 10Ah capacity. Density of anode is 2.2 g/cm3 and density of cathode is 4.5 g/cm3. Report on the battery pack configuration if the required battery pack voltage is 75 volts. If the thickness of anode and cathode is limited to 130 microns (130 x 10-4 cm) calculate the total electrode surface area in each cell. Assume the porosity of electrodes are 30%. Calculate the weight of active materials (anode and cathode) in grams and the total current collector’s and electrolyte membrane areas in (cm2).arrow_forwardDO NOT USE AI NEED HANDWRITTEN SOLUTION Find total impedance of circuit in polar form and power factor.arrow_forward
- Do NOT WANT AI. need diagram fully labeled pleasearrow_forwardCalculate the current magnitude in the coils e1, e2 of theMagnetic circuit, if:ɸa = 3.00 x 10^-3 Wb, φb = 0.80 x 10^-3 Wb, ɸc = 2.20 x 10^-3 Wb L ab = 0.10 m,A ab = 5.0 cm^2L afeb = L acdb = 0.40 mA afeb = A acdb = 20 cm^2 MATERIAL CHARACTERISTICSH (At/m) 240 350 530 1300 5000 9000B (T) 0.7 0.9 1.1 1.3 1.5 1.6arrow_forwardA toroid magnetic circuit is composed of three sections A, B and C, thesection C has an air gap, section A has an 850 round coil thatconsumes a current of 1.2 A. the physical and magnetic properties of each sectionare: Section A: Length = 80 mm, Cross section = 120 mm^2, μr = 400 Section B: Length = 60 mm, Cross section = 40 mm^2, μr = 250 Section C: Length = 50 mm, Cross section = 200 mm^2, μr = 600 Gap: Length = 1 mm, Cross section = 40 mm^2, μr = 1 Calculate:The magnetic field density in each of the sectionsarrow_forward
- 3) Compute the input impedance of Fig. 3. (10 points) Rin R1 R₂ Figure 3 T Vccarrow_forwardShow the workarrow_forward2) A bypass capacitor CE in parallel with RE is added to the above circuit. a) Draw the equivalent small-signal circuit. (10 points) b) Find the input resistance Rib looking into the base. (10 points) c) Find the output resistance looking into the collector, while the source is shorted, i.e. Vs 0 V and Rs = 0 2. (10 points) Vo Vs d) Find the voltage gain A₁ = ✓ using the above equivalent small signal circuit. (10 points)arrow_forward
- EBK ELECTRICAL WIRING RESIDENTIALElectrical EngineeringISBN:9781337516549Author:SimmonsPublisher:CENGAGE LEARNING - CONSIGNMENT
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