Electronics - RL Circuits - Discussion

4. 

If XL= 100 omega.gif and R = 100 omega.gif, then impedance will be

[A]. 141.4 omega.gif
[B]. 14.14 omega.gif
[C]. 100 omega.gif
[D]. 200 omega.gif

Answer: Option A

Explanation:

No answer description available for this question.

Ranjit said: (Sep 2, 2011)  
Z=sqrt(R2 + X2)

Gautham said: (Aug 18, 2012)  
If Z=sqrt(R^2+X^2), then Z=sqrt(100^2+100^2)
Therefore, Z=200. Right?

Chandan said: (Aug 20, 2012)  
z=sqrt(R^2+X^2),then z=sqrt(100^2+100^2)
therefore z=sqrt(20000) so implies sqrt(10000*2)
100sqrt(2) i.e100*1.414=141.4

M.Rizwan Mukati said: (Aug 15, 2014)  
Z = R+jXL.

Then convert it into polar form.

Jyothi said: (Jan 9, 2016)  
R = 100 XL = 100.

Z = sqr(100*100+100*100) = sqr(20000) = 141.4 ohms.

Rajan said: (Nov 16, 2016)  
Another way to express this is due to Resistor and Inductance current flow are not in the same phase they can be shown as below. Total impedance cannot be plus but use trigonometric equation,

Z^2 = Xl^2 R^2.
Z = sqr(Xl^2 R^2).
Z = sqr(100 * 100 + 100 * 100) = sqr(20000) = 141.4 ohms.

Damilare said: (Jul 18, 2017)  
z = sqrt(r squared +XL squared),
Substituting values,we have;
z = sqrt(20000),
z = 141.4ohms.

Gangadharmanu said: (Jan 21, 2018)  
Z =√(r^2*+l^2).
=√(100*100+100*100).
=√(20000).
=141.4.

Swat said: (Aug 21, 2018)  
z= √(r^2+Xl^2).
=√(100^2+100^2).
=√20000).
= 141.4.

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