Adding resistor and capacitor in parallel

AI Thread Summary
The discussion focuses on calculating the total impedance of a circuit with two 8-ohm resistors and a 3-ohm capacitor in parallel. The correct formula for impedance is Z = (1/8 + 1/8 + 1/j3)^(-1), leading to the expected result of 1.44 + j1.92. A participant clarifies that the impedance can also be expressed as 4 ohms and 3j ohms in parallel, resulting in the equation 1/Z = 1/4 + 1/3j. The confusion arises around the numerator 12j, which is derived from multiplying 4 and 3j. The final simplified impedance is confirmed to be Z = 12j/(4 + 3j).
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Hi all,

I am meant to find the total impedance across the circuit but am having trouble with one section of the circuit. This section comprises of 2 resistors and 1 capacitor in parallel.

The two resistors are 8 ohms each and the capacitor is 3 ohms. So the process is meant to be:

Z = ( 1/8 + 1/8 + 1/j3) ^(-1)

and the answer is meant to be: 1.44 + j1.92

I know it is really simple but I can't wrap my head around it at the moment and it's driving me insane! Any help much appreciated.
 
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The secytion can be written as 4 ohms and 3j ohm are in parallel. Then
1/Z = 1/4 + 1/3j
Z = 12j /( 4 + 3j) . Now rationalise the denominator and simplify.
 
Thanks rl.bhat.

I am wondering though how did you get 12j for the numerator? I'm sure it relates to 4 x 3j = 12j somehow but I'm not sure why...
 
When you add 1/4 + 1/3j you get (3j + 4)/12j. This is 1/Z. therefore
Z = 12j/(4 + 3j)
 
Thanks a lot man!
 
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