How to Analyze Complex Circuits

Kirchhoff’s junction rule (or current law) –From conservation of chargeSum of currents entering a junction is equal to sum of currents leaving that junction

Kirchhoff’s loop rule (or voltage law) –From conservation of energy Sum of changes in potential going around a complete circuit loop equals zero

Circuits

What is i through the battery?Label currents. New label for every branch. Pick any arbitrary direction.i through R1 or R4 is same as for batteryCan use loop rule

0412211 =−−− RiRiRiE

+ - + -

- +

Circuits

Equation has too many unknowns so need to apply loop rule again

Take the loop through R2 and R3

0412211 =−−− RiRiRiE

)(

0)(

23

312

22321

321

RRRii

RiRiiiii

+−=

=+−−+=

02233 =+− RiRi

CircuitsNow solve for i1:

041)32(231

11

0412211

=−+

−−

=−−−

RiRR

RRiRi

RiRiRi

E

E

432

321

1

)(R

RRRRR

i+

++

=E

Circuits

What is current Voltage lost (V) in R2 ?Recall that V=iR

222 RiV =

2

41112

412211 0

RRiRii

RiRiRiE

E−+

=

=−−−4

32

321

1

)(R

RRRRR

i+

++

=E

RC - circuit

Circuits where current varies with time

RC series circuit – a resistor and capacitor are in series with a battery and a switch

At t =0 switch is open and capacitor is uncharged so q =0

RC - circuit

Close the switch at point aCharge flows (current) from battery to capacitor, increasing q on plates and V across plates

When VC equal Vbattery flow of charge stops (current is zero) and charge on capacitor is

ECCVq ==

RC - circuit

Want to know how q and V of capacitor and i of the circuit change with time when charging the capacitorApply loop rule, traversing clockwise from battery

0=−−CqiRE

i

RC - circuit

Contains 2 of the variables we want i and qRemember

Substituting gives

0=−−CqiRE i

dtdqi =

E=+Cq

dtdqR

RC - circuit

Need a function which satisfies initial condition q = 0 at t = 0 and final condition of q = C E at t = ∞For charging a capacitor

E=+Cq

dtdqR

( )RCteCq −−= 1E

RC - circuit

Want current as a function of time For charging a capacitor

RCteRdt

dqi −⎟⎠⎞

⎜⎝⎛==

E

( )RCteCq −−= 1E

RC - circuit

Want V across the capacitor as function of timeFor charging a capacitor

( )RCtC e

CqV −−== 1E

( )RCteCq −−= 1E

RC - circuit

Want to know how q of capacitor and i of the circuit change with time when discharging the capacitorAt new time t = 0, throw switch to point b and discharge capacitor through resistor R

i

RC - circuit

Apply the loop rule again but this time no battery

Substituting for i again gives differential equation

i

0=+Cq

dtdqR

0=−−CqiR

RC - circuit

Solution must satisfy initial condition that q0 = CV0

For discharging a capacitor

i0=+

Cq

dtdqR

RCteqq −= 0

RC - circuit

Find i for discharging capacitor with initial condition at i0 = V0/R =q0/RC at t = 0

iRCteqq −= 0

RCteRCq

dtdqi −⎟

⎠⎞

⎜⎝⎛−== 0

Negative sign means charge is decreasing

Circuits

Charging capacitor Discharging capacitor

RCteqq −= 0

RCteRCqi −⎟

⎠⎞

⎜⎝⎛−= 0

( )RCteCq −−= 1E

RCteR

i −⎟⎠⎞

⎜⎝⎛=

E

Define capacitive time constant –greater τ, greater (dis)charging time RC=τ