Relative velocity of A w.r.t. B => A observed from B and B is stationary $\ov{V_{A|B}} =\ov{V_{A}} – \ov{V_{B}}$ ∙Two particles, velocities parallel Same direction: $|V_{A|B}| = v_A – v_B$ in magnitude Opposite direction: $|V_{A|B}| = v_A + v_B$ in magnitude ∙ Two velocities perpendicular $|V_{A|B}| = √{{V^2_A} + {V^2_B}}$ $tanθ = {V_B}/{V_A}$
Displacement (s) = $x_t – x_o$ $d↙{(0 -> t)} = |x_y – x_o|$ Instantaneous velocity = dx/dt Displacement ≦ Distance Velocity ≦ speed a = dv/dt OR a = a ${dv}/{ds}$ Jerk: Rate of change of acceleration w.r.t time $J = {da}/{dt}$ Types of motion and kinematic equations (1) Uniform motion: v is constant and a = 0 s= v x t (2)Uniformly accelerated motion: a os constant and J = 0 V = u + at, s = ut + $1/2$ a $t^2$ $v^2 = u^2 + 2as$ (3)Motion under gravity: g is constant v = u – gt ...
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Magnitude of Resultant: $R = √{P^2 + Q^2 + 2PQ cosθ}$ Direction of Resultant: $tan α = {Q sinθ}/{P + Q cos θ}$ Resultant of two or more forces: $R = √{(ΣF_x)^2 + (ΣF_y)^2} $ Direction of two or more forces: $tan θ = {ΣF_y}/{ΣF_x}$ $ΣF_x$ = Algebric sum of all x components (or x component of resultant) $ΣF_y$ = Algebric sum of all y components (or y component of resultant) θ = Angle of ‘R’ with x-axis When an object is inclined on a plane at angle θ Component along plane -> mg sinθ Component perpendicular to plane -> ...
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AND Gate Y = A.B OR Gate Y = A + B NAND Gate Y = $\ov{A.B}$ NOR Gate Y = $\ov{A + B}$ EX-OR Gate Y = A ⊕ B EX-NOR Gate Y = $\ov{A⊕B}$ $\table Name,\text"Statement of the law"; \text"Commutative Law",A.B = B.A;,A+B=B+A;\text"Associative Law",(A.B).C=A.(B.C);,(A+B)+C=A+(B+C);\text"Distributive Law",A.(B+C) = AB+ AC; \text"AND Laws",A.0=0;,A.1=A;,A.A=A;,A.\ov{A}=0;\text"OR Laws",A+0=A;,A+1 = 1;,A+A=A;,A+\ov{A}=1;\text"Inversion Law",\ov{\ov{A}}=A;\text"Other Important Laws",A+BC=(A+B)(A+C);,\ov{A} + AB = \ov{A} + B;,\ov{A} + A \ov{B} = \ov{A} + \ov{B};,A+AB = A;,A+ \ov{A}B = A +B$ De-Morgan’s Theorem $\ov{AB} = \ov{A} + \ov{B}$ $\ov{A+B} = \ov{A} . \ov{B}$
$f_o = 1/{T_c + T_d}$ $T_{on} = T_c = 0.693 (R_A + R_B)C$ $T_{off} = T_d = 0.693 R_B C$ $f = {1.44}/{(R_A + 2 R_B) C}$ Duty Cycle, $D = {R_A + R_B}/{R_A + 2 R_B} $ Load Regulation, $%L.R. = {V_{NL} – V{FL}}/{V_{FL}}$ x 100
$α_{dc} = {β_{dc}}/{a + β_{dc}}$ $β = {α}/{1 – α}$ $β_{dc} = I_C/I_B$ $I_{CEO} = (1 + β_{dc}) I_{CBO}$ $I_{CEO}$ = reverse saturation current $I_E = I_B + I_C$ $γ = I_E/I_B = (1 + β_{dc})$ Current gain, $A_I = I_o/I_i$ Voltage gain, $A_v = V_o/V_i$ Power gain, $A_p = A_v$ x $A_I$ $V_i = {R_i}/{R_i + R_s}$ x $V_s$
$\table Parameter,HWR,FWR,Bridge-rectifier;\text"Avg. load current "(I_{Ldc}),I_m/π, {2I_m}/{π},{2I_m}/{π};\text"Maximum avg. load voltage" (V_{Ldc}), V_m/π, {2V_{m}}/{π},{2V_{m}}/{π};\text"RMS load current"(I_Lrms), I_m/2,I_m/√2 , I_m/√2; \text"RMS load voltage" (V_{Lrm}), V_m/2, V_m/√2,V_m/√2;\text"DC load power" P_{dc}, {I^2_{m}}/{π^2} R_L,{4I^2_{m} R_L}/{π^2},{4I^2_{m} R_L}/{π^2};\text"Max. rectification efficiency (η)",40%,81.2%,81.2%;TUF,28.7%,69.3%,81.2%;\text"Ripple factor",121%,48%,48%;\text"Ripple Effect",50 Hz, 100 Hz, 100 Hz;\text"Number of diodes used", One, Two, Four;\text"Center tap transformer",\text"Not req.",\text"Very much req.",\text"Not req.";\text"Transformer core saturation",Possible,Not-possible,Not-possible;PIV,V_m,2V_m,V_m;\text"Expression for peak load current",I_m={V_m}/{R_S + R_F + R_L},I_m={V_m}/{R_S + R_F + R_L},I_m={V_m}/{R_S + R_F + R_L}$ Ripple Factor, $RF = {V_{RMS}}/{V_{Ldc}} = {1}/{4√3 fCR }$ …For full wave or bridge rectifier circuit RF for HWR = $1/{2√3 fCR}$ Load current, $I_L = V_z/R_L$ $I_z$ -> zener ...
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∑B.S. – ∑F.S. = Last R.L. – First R.L. = ∑Rise – ∑Fall A(B.S.) = 1.250 -> B(F.S.) = 2.250 There is a fall from A to B Difference in R.L. = Difference in staff reading Note: There is no entry in the first row of rise and fall column. Amount of fall = Following staff reading – Preceeding staff reading Rise = B.S. – F.S. R.L. = R.L.(previous) – Fall R.L. = R.L.(Previous) + Rise (Previous) I.S. – I.S. = Rise/Fall H.I. = R.L. + B.S. (Previous) H.I. = R.L. + I.S. I.S. – Fall = B.S. Collimation plane ...
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KCL : ΣI = 0 KVL: ΣV = 0 Superposition theorem: $I_L = I’_L + I”_L$ No specific formulae in Thevenin’s theorem