Engineering Finding R and C from a parallel RC circuit

AI Thread Summary
To find the resistance (R) and capacitance (C) values in a parallel RC circuit with an impedance of Z = 105000 Ω ∠ -27 degrees at 50 Hz, the impedance can be converted to rectangular form as Z = 9355.69 - j47669. The relationship for parallel RC circuits is given by 1/Z = 1/R + jωC, where ω is the angular frequency. By calculating 1/Z and expressing it in a + jb form, the real part corresponds to the conductance (1/R) and the imaginary part relates to the capacitive reactance (ωC). Equating these components will yield the values for R and C, with scientific notation recommended for clarity.
Agent47
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Homework Statement



A RC parallel circuit with an impedance of Z = 105000 Ω ∠ -27 and frequency of 50 Hz.

Homework Equations



I need to find out what the R and C values using the information given.

The Attempt at a Solution



I know the impedance for a parallel RC can be calculated from

1/Z = 1/R +jwc

Z can be converted to Z = 9355.69 -j47669 but from there i am completely confused on what to do next.

Any help will be much appreciated.
 
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Well you can calculate 1/Z and express it in a+jb form...Compare the real and imaginary parts then.
 
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If i do 1/z i would get,

0.000008485 +j0.000004323

but I am still not sure where R and C would come from.
 
Agent47 said:
I know the impedance for a parallel RC can be calculated from

1/Z = 1/R +jwc

Agent47 said:
If i do 1/z i would get,

0.000008485 +j0.000004323

but I am still not sure where R and C would come from.

[scientific notation would help curtail all those zeros :smile:]

Equate the terms of your expression above to the numerical values of the conductance components...
 
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I want to thank you both for your help.
 

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