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- From step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?(inference regarding signals from pole-zero information) Handwrite and step by step solutionsQ2/Find (1101100)2 + (1100111)Ex-3 - (1011011)Gray= ( )Gray O 1010010 O 1100101 O None of them O 1111111 O 1000010 101100011
- io2 (in ampere) due to 12 only= Oa. 0.9277621122789 Ob. -4.6388105613945 Oc. 2.3194052806973 Od. -2.3194052806973 (Note that io=i01+io2+io3, where io1 is due to V1, io2 is due to 12 and io3 is due to I1) io3 (in ampere) due to 11 only= Oa. 0.14201486798257 Ob. 1.4201486798257 Oc. 0.71007433991284 Od. -0.21302230197385 (Note that io=i01+io2+io3, where io1 is due to V1, io2 is due to 12 and io3 is due to I1) io (in ampere) = (Note that io=i01+io2+io3, where io1 is due to V1, io2 is due to 12 and io3 is due to I1) Power dissipated by R2 (in watt) =The true value of a current is 10mA and it is indicated with a digital ammeter as 12mA. The absolute error for this measurement system is _mĀA. O a. 0.16 O b. 0.2 O C. -2 O d. 2How to solve 240 (240/3.15/-31.4) the same way as 120(5.722 /-65.2) equated to 288.01-j623.32
- التاریخ | موضوع الدرس Vincoms) -220v AL=10k C = 100 ME e The FwRcct. %3D Find: 1. Dcvaltage and Current AS/p=1/5 ii the ripple Ualtage Tit ripple factorso then how is Rth 0.8695 kOhms in the original explanation?ALL FINAL ANSWER SHOULD BE ACCURATE TO 3 DECIMAL PLACES USE POSITIVE ANGLE EQUIVALENT, NOT GREATER THAN OR EQUAL TO 360 DEGREES "If Vb is equal to 1274308 and Vbc is 219.972338, Vab should be
- For which measurement is this oscilloscope time difference reading useful? Oscilloscope-XSC1 Time Channel_A Channel B 89.282 V -91.176 V 250.000 us 169.064 V Reverse -168.880 V -337.944 V 753.788 us T2-T1 503. 788 us -180.458 V Save Ext. trigger Timebase Channel A Channel B Trigger Scale: 200 us/Div Scale: 100 V/Div X pos. (Div): 0 Y/T Add B/A A/B 100 V/Div Edge: F2 AB Ext Scale: Y pos. (Div): 0 Y pos. (Div): 0 Level: AC DC AC 0 DC O Single Normal Auto None SER RI MEc) Given the changed lines of code below, answer the following questions. ANALOG_COMPARE_SETUP: LDI R16,0 STS ADCSRA, R16 LDI R16, 0b0e*** *** STS ADCSRB, R16 LDI R16,0 STS ADMUX, R16 LDI R16,0b00000000 OUT ACSR, R16 RET What signals are now being compared? Determine if a comparator interrupt will occur (show work): Time (s) AINO (volts) AIN1 (volts) 0 1 4 1 5 4 2 1 4 3 1 0 4 1 4 4 0 alvil 5 6 5 5 Interrupt (y/n) n/a initial stateVth, Zth and In are needed for Vth the answer is 3.47 Angle 123 but it not sure so pls answer the way you like! Please use three decimal places, thank you!