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jas_singh

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Hi

I have 5 questions with answers with are:

1) A resistor of 28_ is connected to a inductor of 12mH and a capacitor of 24μF. Calculate

the impedance presented to the 50Hz supply.

XL = 2πfL = 3.77_, XC = 1 / (2πfC) = 132.63_, Z = √ (R2 + (XC- XL)2 = 131.87_

2) A resistor of 38_ is connected to a inductor of 124mH and a capacitor of 100μF.

Calculate the impedance presented to the 50Hz supply.

XL = 2πfL = 38.96_, XC = 1 / (2πfC) = 31.83_, Z = √ (R2 + (XL- XC)2 = 38.66_

3) A resistor of 12k_ is connected to a inductor of 56.92mH and a capacitor of 178μF.

Calculate the impedance presented to the 50Hz supply.

XL = 2πfL = 17.88_, XC = 1 / (2πfC) = 17.88_, In this case XL=XC, so Z=R = 12k_

4) A resistor of 164_ is connected to a inductor of 84mH and a capacitor of 18μF. Calculate

the impedance presented to the 50Hz supply.

XL = 2πfL = 26.39_, XC = 1 / (2πfC) = 176.84_, Z = √ (R2 + (XC- XL)2 = 222.56_

5) A resistor of 216_ is connected to a inductor of 136mH and a capacitor of 124μF.

XL = 2πfL = 42.73_, XC = 1 / (2πfC) = 25.67_, Z = √ (R2 + (XL- XC)2 = 216.67_

Does anyone know how to work them out as im having problems getting the correct answer.

thanks

 
Hi 1) A resistor of 28_ is connected to a inductor of 12mH and a capacitor of 24μF. Calculate

the impedance presented to the 50Hz supply.

XL = 2πfL = 3.77_, XC = 1 / (2πfC) = 132.63_, Z = √ (R2 + (XC- XL)2 = 131.87_
Is that not the correct answer?

 
yes thats the correct answer but not sure how they got it

do you know how to do the first and show working out??

 
Reactance of inductor XL = 2 x 3.141 (pi) x 50 x .012 = 3.769R

Reactance of Capacitor XC = 1/2 x 3.141 x 50 x .000024 = 132.65R

Circuit reactance = XC - XL = 128.88R

Z = square root of (28 x 28 + 128.88 x 128.88) = sq root of 17394.05 = 131.88R

(Inductor has to be in Henry's for calculation and Capacitor has to be in Farads)

Any help?

 
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