ELECTRONICS

A loop that does not contain other loops

Sum of all current leading to a node = 0 (conservation of charge( so if all currents facing the node - the sum = 0 if some facing and some moving away then can rearrange to that sum of those moving away equal those going towards

Charge is in coulombs - 1.6E-19 Voltage is volts = 1 J/C Current is I in amps - its C/s Power is Watts and P = IV (also I^2 /R or V^2/I (and of course V = IR

resisitivity is an inherent property of a material (unit is ohm m) - Resistance as a whole depends on geometry of as well as resistivity So R = p *(L/A)

Inverse of Resistance (1/R) - its G

W = IV so 1000 = 120 * I I = 8.33 120 = 8.33 * R R = 14.4

Idea lvoltage source- Output voltage doesnt change - 0 internal resistance - Current changes based on load Ideal current source - Load doesn’t effect output current (voltage changes based on load) and infinite internal resistance

branch a portion of circuit with 2 terminals Node - a junction of 2 or more branches

Sum of the voltages around a closed loop is 0 eg so if have 2 sources their sum = 0

So in series I is the same and V is the sum of V’s R is the sum of the R’s In parallel I is the sum, V is the same and R is(1/R = 1/R +1 /R + 1/R)(so int his case - conductivity is summed G is sum of G’s)

Basicaly a bunch of things in series - if we have a bunch of resistors in series its a voltage divider - If there are a bunch of resistors in a row the voltage we measure can vary based on how much of the resistors we are measuring across

A bunch of resistors in parallel

Has internal resistance or impedance But basically if the internal resistance is «< Rl (resistance of load) can essentially say that the vl = vs (voltage across load equals that of V source

internal imedance infinity - So the OUTPUT current depends on R - look at derivation for that - KEY HERE IN DIAGRAM THEN - our internal resistance is PARALLEL to our load resistance So we see the opposite relationship where rs»> Rl then Is = Il(go through derivation)

You want to do it in parallel with a resistor (in parallel voltage remains the same) However to minimize loading error you want the DMM to have Very high resistance such that the R equivalent for this parallel part essentially just equals the resistance of the resistor in circuit (DO OUT THE math - For our parallel section with the DMM the Req is 1/Req = 1/Rc + 1/Rdmm Which turns into :Rc*Rdmm / (rc + RDmm) SO if Rdmm»_space;> Rc then it just turns into Rc which is what we want - reflects the circuit

We want it in series with minimal resistance (so again thinking of resistance - in series add them up - soto make the equivalent resistance = the resistance in circuit want Rc»»> RDmm

Essnetiall % error for V (so V load - V battery / V battery)*100 OR for REsistor ITS different -Rt / (Rt +Rm) Note m is the load or meter and T is the soruce NOTE always negative error because ITS always less than NOTE -larger DMM resistance makes or error smaller

% error for current again and for Resistance Er = - Rm/(Rn + RM) x 100 RM being fo the load I belive LOADING ERROR DECREASES WITH SMALLER DMM RESISTANCE which we showed

Draw current direction Define clock wise or countercloskwise to sum voltages (use the same convention in every loop) - APPLY THE KVL (V in series should add up to 0 (or to the battery amount) Or can apply kirchoff current law (at a node the sum of current s= 0

Basically any network of sources and resistors can be seen as a voltage source and resistor in series (Vt is voltage source) and RT is resistance of netowrk

Same idea as thevening but a CURRENT source and a parallel resistor - r = Rn

Rt = Rn

Take away voltage source replace with wire (note we’re calculating resistance Load sees) - so then we turn the remaining resistors into one equivalent resistor in regard to the load (in Voout or a b