Posted by Rony on Thursday, February 18, 2010 at 12:56pm.
First get the effective resistance of R1 and R2 in series. It is 24 ohms. That in parallel with 8 ohms has an overall circuit resistance Reff geiven by
1/Reff = 1/24 + 1/8 = 4/24 = 1/6
Therefore Reff = 6 ohms.
The total current going through the circuit (battery) is
I = V/Reff = 1.0 amperes.
The total current going through the R1 and R2 is V/(R1+R2) = 6/24 = 0.25 A
The potential drop across R2 is
0.25A*9 ohms = 2.25 V
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