gandanmo2
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When a d.c. supply at 240 V is applied to the ends of a certain
choking coil, the current in the coil is 20 A. If an a.c. supply at 240 V,
50 Hz, is applied to the coil, the current in the coil is 12.15 A.
Calculate the resistance, impedance, and inductance of the coil.

now to find resistance its just v=ir right? to get 12ohms
and impedance is z=v/i? or am i wrong? it seems to easy. any help please?
thanks so much!
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uberteknik
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(Original post by gandanmo2)
When a d.c. supply at 240 V is applied to the ends of a certain
choking coil, the current in the coil is 20 A. If an a.c. supply at 240 V,
50 Hz, is applied to the coil, the current in the coil is 12.15 A.
Calculate the resistance, impedance, and inductance of the coil.

now to find resistance its just v=ir right? to get 12ohms
Yes. d.c. resistance is

R = \frac{v}{I} = \frac{240}{20} = 12\Omega

(Original post by gandanmo2)
and impedance is z=v/i? or am i wrong? it seems to easy. any help please?
thanks so much!
Yes, but remember the question is a bit confusing since both d.c. and a.c. are stated as 240V in this question.

The resistance uses the d.c. voltage value.

The impedance uses the a.c. voltage value.


Impedance is frequency dependent, hence the 50Hz is brought into play when calculating the coil inductance.

Do you know the equation for the inductive reactance of a coil?
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gandanmo2
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(Original post by uberteknik)
Yes. d.c. resistance is

R = \frac{v}{I} = \frac{240}{20} = 12\Omega



Yes, but remember the question is a bit confusing since both d.c. and a.c. are stated as 240V in this question.

The resistance uses the d.c. voltage value.

The impedance uses the a.c. voltage value.


Impedance is frequency dependent, hence the 50Hz is brought into play when calculating the coil inductance.

Do you know the equation for the inductive reactance of a coil?
AHHH! i see i kool so then youd get r= 240/12.15
and then sub that into z(squared)= r^2 + xc^2 then rearrange that to get l!? right?
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gandanmo2
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(Original post by uberteknik)
Yes. d.c. resistance is

R = \frac{v}{I} = \frac{240}{20} = 12\Omega



Yes, but remember the question is a bit confusing since both d.c. and a.c. are stated as 240V in this question.

The resistance uses the d.c. voltage value.

The impedance uses the a.c. voltage value.


Impedance is frequency dependent, hence the 50Hz is brought into play when calculating the coil inductance.

Do you know the equation for the inductive reactance of a coil?
Also! theres this question thats completely mind fed me!

A 230 V motor has an armature circuit resistance of 0.6  . If the
full-load armature current is 30 A and the no-load armature current is
4 A, find the change in the back e.m.f. from no-load to full-load.
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uberteknik
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(Original post by gandanmo2)
AHHH! i see i kool so then youd get r= 240/12.15
and then sub that into z(squared)= r^2 + xc^2 then rearrange that to get l!? right?
That gives the magnitude only.

Resistance is the ratio between voltage and current for d.c. conditions. i.e. Ohms law is magnitude only.

Reactance is resistance measured at a specific frequency. Hence XL but still only magnitude.

Impedance introduces the frequency dependent magnitude and phase relationship between voltage and current of the inductor.




I = \frac{E}{X_L}

X_L = \frac{240}{12.15}

also

X_L = \omega L

Substitute and rearrange:

X_L = 2\pi fL

L = \frac{X_L}{2\pi f}


Z = j\omega L

Z = R \ + \ jX_L


Z^2 = R^2 \ + \ X_L^2

\lvert Z \rvert = \sqrt\big(R^2 \ + \ X_L^2 \big)

don't forget to state the argument

\mathrm{arg \ }Z = \phi = tan^{-1}\big(\frac{X_L}{R}\big)

Do you know how to state the impedance in exponential form?
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gandanmo2
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(Original post by uberteknik)
That gives the magnitude only.

Resistance is the ratio between voltage and current for d.c. conditions. i.e. Ohms law is magnitude only.

Reactance is resistance measured at a specific frequency. Hence XL but still only magnitude.

Impedance introduces the frequency dependent magnitude and phase relationship between voltage and current of the inductor.




I = \frac{E}{X_L}

X_L = \frac{240}{12.15}

also

X_L = \omega L

Substitute and rearrange:

X_L = 2\pi fL

L = \frac{X_L}{2\pi f}


Z = j\omega L

Z = R \ + \ jX_L


Z^2 = R^2 \ + \ X_L^2

\lvert Z \rvert = \sqrt\big(R^2 \ + \ X_L^2 \big)

don't forget to state the argument

\mathrm{arg \ }Z = \phi = tan^{-1}\big(\frac{X_L}{R}\big)

Do you know how to state the impedance in exponential form?
aah i got it thanks so much man!!!
also if you could help me with this one it would be appreciated!

A 230 V motor has an armature circuit resistance of 0.6  . If the
full-load armature current is 30 A and the no-load armature current is
4 A, find the change in the back e.m.f. from no-load to full-load.
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