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					Circuit Basics
 Ampere’s LawI(t)= dq/dt
Current = i = 1 C/s
Charge = q = C
Charge 1 electron = -1.602x10^-19C
1 C = 6.24x10^18
volt = joules/coulomb = dw/dq
p(t) = dw/dt = dw/dq*dq/dt = v*i
Watts= Volts*Amps
p>0 absorbed, p<0 delivered
Passive sign convention – current runs from + to –
Voltmeter measures across element(open circuit)
Ammeter measures inline (thru) element(Short circuit)

Circuit Elements
Linear elements hold 2 properties:
1.Superposition: v1v1, i2v2, theni1+i2v1+v2
2.Homogeneity: iv, then kikv
Active elements are capable of supplying energy( w=∫(vi,τ,-∞,t)< 0)
Passive elements absorb energy( w=∫(vi,τ,-∞,t)≥ 0)
Resistivity: ohms-cm(ρ)(conductors=low, insulators=high)
Resistance: impede the flow of current; R (Ω): R= ρL/A
Ohm’s Law: v=iR ;
Conductance: G; Siemens(S); G=1/R (S=1/ Ω); i=Gv
Resistance changes with temperature, resistance can reach 0; superconductor
Home wiring(12 gague) is rated 20-26A
Resistor values are identifiable by the 4 bands(1st and 2nd digits, multiplier and tolerance)
Power in a resistor: p=vi=(iR)i=i^2Rvi=v(v/R)=v^2/R; Power is positive(absorbed)
Energy in a resistor: w=∫(p,τ,-∞,t)= ∫(Ri^2,τ,-∞,t) = R∫(i^2,τ,-∞,t);
Since i^2 always positive energy w is always positive and resistor is a Passive element
Short circuit 0 Ω resistance; open circuit ∞ Ω resistance
Source: a v or i generator capable of supplying energy to a circuit
Independent source: v or i generator not dependent on circuit variables
Ideal source is a v or I generator independent of i thru v source or v thru i source(circle)
Switches: closed(v=0), open(i=0); types: spst:single pole single throw; spdt:double throw
Spdt Break before Make and Make before Break
Dependent sources types: ccvs, vcvs, vccs, and cccs
Voltage Controlled Voltage Source: b=gain b: volts/volt
Voltage Controlled Current Source: g=gain g: amperes/volt
Current Controlled Current Source: d=gain g: amperes/ampere
Current Controlled Voltage Source: r=gain g: volts /ampere

Resistive Circuits
3 laws of Resistive Circuits: Ohm’s Law, KCL & KVL
Node :junction of 2 or more elements
Loop: Closed path around a circuit
KCL: the sum of all currents entering a node is 0(current in =Current out)
KVL: The sum of voltages around a loop is 0( from + to -  pos, - to +  neg)
Series Resistors: Voltage divider circuit: vn=Rn/(R1+R2+…Rn)*vs
Equivalent Resistance: Rs=R1+R2+…Rn); i=vs/Rs;
        Source must drive the same current or deliver the same power P=Vsi=vs^2/Rs
Parallel Resistive circuits with a voltage source: v1=v2=…vn
Parallel circuits w/ a current source: KCL: is=i1+i2; i1=v/R1
        is=v/R1+v/R2= R1+R2/(R1*R2)*v v=R1R2/(R1+R2)*is
        i1=v/R1=R2/(R1+R2)*is ; i2=v/R2=R1/(R1+R2)*is
Equivalent Resistance: Gp=1/Rp=(1/R1+1/R2+…1/Rn)
Source sums: va +/- vb in series; ia +/- ib in parallel;
        No va+vb in parallel(unless va=vb); No ia+ib in series (unless ia=ib)

Resistive Circuit Analysis
Node Voltage and Mesh current
Node voltage: use N-1 KCL equations
Vac=Va; Vbc=Vb; Vab=Va-Vb
i2=va/R2; i3=vb/r3, i1=Vab/r1=(va-vb)/R1
KCL Equations of nodes in form: is=G1va+G2(va-vb)+G3(va-vc)
Rearrange equations to this form: is= (G1+G2+G3)va - G2vb - G3vc
Solve matrix equations
Nodal analysis with V and I sources: 3 scenaios
     1. Voltage source connects to one node and Ground
     2. voltage source has a Series Resistor
     3. Voltage source lies between 2 non-reference nodes
1. since va=Vs, only need to analyze other nodes using KCL
2. 2 solution methods:
     1. create a new node, isolating Vs in a node-ground configuration
     2. transform Vs and series resistor to current source with parallel resistor
     3. Supernode method: KCL the supernode and substitute va-vb=vs to solve
If the circuit contains a dependent source, it must be supplemented with an addtl eq.
Mesh current: KVL

				
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posted:9/22/2011
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