The switch has been closed for a long time for t < 0. Find i(t) for t > 0. 5Ω www i t=0 1.5 H m 10 22 9 A
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- t of For R1-47 0, R2=72 Q, R3=104 Q, V1=6 V, 1=40 A & K1=4.5, use the principle of superposition to find : R2 KIV, R3 V1 I1 R1 V01 (in volt) due to V1 only= a. 0 b. -0.028190603132289 C. -0.018793735421526 d. -0.0093968677107631io2 (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) =(i) Convert (101011000.1101)2 to Hexadecimal (ii) Using 1’s complement, perform subtraction X-Y and Y-X Where X= 1010100 Y=1000011
- %VY l1, M O 1::18 And----- are two types of FA- * with output DFA and NFA Moore and Mealy machine FA and Regular expression TG and DFA O we cannot convert Moore machine to the * Mealy machine True False The following pairs of regular expression * define the the same language (ab) * and a*b* (ab)*a and a(ba)* O a (ba+a)* and aa*b (aa)*(E+a) and aa* II3/ 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 .Q2/Find ((90)10/(36)8)* (F1)16+ (11ΟΟ111)2=( )Gray O 110100111 O 100101100 O 1000101100 O 1010100111 None of them Ο Ο Ο Ο Ο
- 85[1+0.4432/1-.85(sqrt(48.2625/V))]=100 Solve for V(inference regarding signals from pole-zero information) Handwrite and step by step solutionsThank you for your help so far. The original question from my homework is the 1st attached image, but I just want to know 2 things. 1) How to calculate iD av and iD max? And if you're feeling extra nice; 2) What are iD av and iD max called on a datasheet? I'm looking at the data sheet and I dont see any currents related to "D". I attached a 2nd picture of the datasheet I'm looking at. If the 2nd question is too much for 1 solution on this site, please just answer my 1st question and I'll follow up with my 2nd. I don't want to type this whole question again, lol. Thanks again!
- If the diagram shown in Figure 9-6 represents a three-bit Gray code counter, what binary value would exist at 'X'? Present State Next State Q2 Q1 Q0 Q2 Q1 Q0 000 1 (100 001) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (111) (010) 1 1 1 A 110 101 C 010 011c) 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 stateRouth's method table