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leelu leelu
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10 years ago
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R1 = 600 ohms
R2 = 400 ohms
R3 = 560 ohms
R4 = 800 ohms

I freeze when it is combined series and parallel circuit problems, all I know what to do is adding R1 and R2 as they are parallel. If anyone can show me a list of steps to solve the problem I would greatly appreciate it.
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o2bleao2blea
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10 years ago
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10 years ago
What we need to know is the resistance the source "sees."  At the terminals where the source connects to the resistor network, the equivalent resistance is:
r4 in parallel with [ (r1 in parallel with r2) + r3] = r4 || [(r1 || r2) + r3]
r1||r2 = 400*600/[400+600] = 240ohms
= 800 || [ 240 + 560] = 800 || 800 = 400ohms
So the current leaving the source I = 12/400  = 30ma.  At the point where r4 connects, half goes thru r4, half the other path since 800 ohms is seen in both paths.  We know the Vr1 = Vr2, the voltage across r1 = the voltage across r2 because end of one is connected to the end of the other (they are in parallel).  Thus we can write
600*i1 = 400*i2 =>600/400 = 1.5 => 1.5*i1 = i2  AND i1 + i2 = I/2 => substitute in 1.5i1 for i2 => 2.5*i1 = I/2
i1 = I/5 = 0.2*I = 6ma => i2 = I/2-I/5 = 3*I/10 = 0.3*I = 9ma
The current thru r4 is the same as the current thru r3 since 800 ohms is seen down both paths.  
Vr4 = 15*0.8 = 12V
Vr3 = 15*0.56 = 8.4V
Vr1 = 6*0.6 = 3.6V
Vr2 = 9*0.4 = 3.6V
Notice that 8.4+3.6 = 12 so it checks
Hope this helps.
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