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# Magnitude of Impedience

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See attached dwg. Can you please show me how to find the magnitude of Z (The impedience "looking in") as a function of frequency from 10Hz to 100KHZ? Note the inductor is 1mH the resistor 100 Ohms and the cap 1uF.

https://brainmass.com/engineering/electrical-engineering/magnitude-impedance-72183

## SOLUTION This solution is FREE courtesy of BrainMass!

Theory and Approach :

1 . Looking at the figure below, RC form a parallel circuit i.e. current flowing into the inductor is divided at the RC junction. This also implies that if you add up the total current flowing into R and C, you will get the current flowing in the inductor.

So we can conclude that impedance due to R and C are in parallel and which in tern in series with impedance due to the inductor.

2 . Impedance of a resistor is called "resistance" = XR = R
Impedance of a capacitor is called "capacitive reactance " = XC = 1/jwC
Impedance of a inductor is called "inductive reluctance " = XR = jwL

Problem solving

Z = XL + (XR || XC) = jwL + (R || 1/jwC) = jwL * (1 + jwRC) / jwC = L/C (1 + jwRC) -------- ( 1)

Aside:
R|| 1/jwC = (1+ jwRC) /jwC ------- (A)
Use equation (A) in (1)

Substitute values for
L = 0.1 mH
R = 100 ohm
C = 1uF
and w = 2* pi* f =
i) When f= 10 ohm; w = 2 * pi * 10 = 20 * pi
ii) When f = 100K ohm; w = 2 * pi * 100K = 200 K * pi

Doing the math

Substituting the numbers in equation (1) we have
Z = L/C (1 + jwRC)

Case 1, when f = 10 ohm ;
Z = 0.1 K (1.06) = 0.106K ohm = 106 ohm

Case 2 when f = 100 K ohm
Z = 0.1K (629.4) = 62.9k ohm

Aside :
1) L/C = 0.1 m/ 1 u = 0.1 K
2) RC = 100 u

PS: pls verify the numbers with a calculator, I have done some hand calculations.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!