Calculating ratio of voltage help

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To calculate the voltage across a 4.7k ohm resistor and a 100nF capacitor at a frequency of 10 Hz, the correct approach involves using the impedance formula for each component. The total impedance combines the resistor and capacitor impedances as vectors since they are at right angles. For the resistor, the formula Vr/v = R/z applies, while for the capacitor, Vc/v = (1/wC)/z is used. It is crucial to clarify mathematical expressions with parentheses to ensure proper operator precedence. Understanding the circuit configuration and the power source is essential for accurate calculations.
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Homework Statement



At the given 10 Hz frequency calculate the magnitude of the voltage across a:

(1)4.7kohm resistor
(2)100nF capacitor

Homework Equations



Vr/v=R/z=R/sqrtR^2+(1/wC)

Vc/v=(1/wC)/z=(1/wC)//sqrtR^2+(1/wC)

The Attempt at a Solution


(1)4.7/sqrt(4.7^2+2*PI*4.7^2) ------i am not sure if this correct

and for part 2

(2)1/(2*PI*100*10^-9)/sqrt()

not sure of what value to use for the R
 
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There should be more information. What information was given about the power source? What circuit was described?
 
oh it had 10V power source
 
i think i have to calculate the voltage across each component as they are in series but not sure how to use impedance on this question
 
It appears that you are using as the formula for the total impedance:

z=sqrtR^2+(1/wC)

You have not derived this correctly.

You are aware that the total impedance is the sum of the impedance of R + the impedance of C. But because these individual impedances are at right angles to each other, you must add them like vectors at right angles to each other.

Complete this: the impedance of a capacitor of 100nF at 10Hz =
 
ok z=1/2PI*10*100*10^-9 then
 
joe007 said:
ok z=1/2PI*10*100*10^-9 then

Consider using parentheses to clarify the mathematical operator precedence in your expressions. The X2 and X2 icons in the post composition header let's you create superscripts and subscripts, too. Thus:

Zc = 1/(2*pi*10*100*10-9)
 
Now, basing your calculations on a right-angled triangle, add the two series impedances.
 
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