18 USE SOURCE TRANSFORMATION TO FIND Io 3kn Io 6kn To= -/5r %3D 4mA
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- 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) =INFORMATION THEORY No wrong answer plz*85[1+0.4432/1-.85(sqrt(48.2625/V))]=100 Solve for V
- (i) Convert (101011000.1101)2 to Hexadecimal (ii) Using 1’s complement, perform subtraction X-Y and Y-X Where X= 1010100 Y=1000011subject DLD plzz solve urgent with detail thnx3/ A Smufodail signal shown in figure (1) be low: Time (t) VPPニRY Upp is 12v and T= 0.08 Sec I- Explain in details the method of Converting the above signal to a start with otsitel bits stream 0oo0 upto 1111 I1- what is the bit reate for the process a buve ? I7l- ealeulate: Signel Signl to noi se ratio and noise power .
- Solve for the output of the full adder showing Boolean values at all labeled wire locations. Convert to decimal answer. P=d9 Q=d5 CI=d1b) Given the changed lines of code below, answer the following questions. ANALOG_COMPARE_SETUP: LDI R16,0 STS ADCSRA, R16 T LDI R16,0b00000000 STS ADCSRB, R16 LDI R16,0 STS ADMUX, R16 LDI R16,0b00001000 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) 1 4 5 4 1 4 1 0 1 4 5 4 5 0 0123456 Interrupt (y/n) n/a initial stateQ1/The Gray code ( 10110111101)= ( ) in decimal number O 1749 O 1877 O 1750 1753 869 None of them 931 867 Q3/Find the solution of the * following operation for octal number 717*0.33