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AUTOMOTIVE ELECTRONICS:
(LAB EXPERIMENTS ON BOTTOM OF THIS SHEET.)
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LECTURE #4

SERIES-PARALLEL RESISTORS CURCUITS:  Connect one end of  2 small light bulbs together.  Then connect one end of 2 20ohm bulbs together.  Now connect the free ends of each pair to the battery plus & minus prospectively.  The entire battery voltage drops across each (series) resistor pair, but the current is divided in proportion to the (series) resistor pair sizes.  The pair with the less resistance allows more current to flow!!!  See webpage problem.
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Example4a:   The 1st (series) resistor pair is r1+r2, 10+10 ohms = 20ohms.  The 2nd is r2+r4,  20+20=40ohms.  The total resistance is. . .

R=(20x40)(20+40) = 800/60 = 13.3333...ohms
I = E/R = 12volts/13.333ohms = .9amps 

1=60/60  Since its indirectly proportionate, we must subtract its proportion from 1.
60/60 - 20/60ths = 40/60ths of  = (2/3) times .9amps = .6amps  though r1+r2.
60/60 - 40/60ths = 20/60ths of  = (1/3) times .9amps = .3amps through r3+r4.
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Example4b:   The 1st (series) resistor pair is r1+r2, 20+10 ohms = ____ohms.  The 2nd is r2+r4,  20+40=_____ohms.  The voltage is 60volts.  
Find the total resistance is. 
Find the currents in each branch.  Ix = ______amps?  Iy = _____amps
Find the total current.  It = _________amps
Show your work by typing you calculations in this space and emailing your homework to JML Online.
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