# hnc question 4 (last one i hope!!!)

Discussion in 'Homework Help' started by redshaw, Aug 4, 2008.

1. ### redshaw Thread Starter Member

Jul 15, 2008
12
0
a coil of resistance 8 ohms and inductance 150 mH is connected in series with a 100uF capacitor across a 240V 50 Hz ac supply. calculate a.) the circuit current, b.) the circuit phase angle, c.) the p.d across the coil, d.) the p.d across the capacitor and e.) the power dissipated.

i have calculated a.)

Xc=1/2piefc
Xc=1/2pie*50*100*10^-6
Xc=31.83 ohms

Xl=2piefl
Xl=2pie 50*150*10^-3
Xl=47.12 ohms

Z=square root r^2 + (Xl-Xc)^2
Z=17.25 ohms

I=V/Z
I=240/18.37
I=13.9 A

b.) as

tan theta=47.12-31.83/8
tan thata=1.911
theta=arctan 1.911
theta= 62.38 degree

for c and d

is pd phase difference, potential drop, power dissipated - any ideas

i have read section in principles of elec technology (cotton) about phase difference but am struggling to understand it and put it into place with the figures/data i have.

e1=OPsinwt
e2=OQsin(wt-theta)

did consider Ur=IR for the couil but was unsure aboutcapacitor

any help/formulas would be welcome

d.)

P=I^2/R
P=13.9^2/8
P=24.15 W

hopefully the rest of my working is correct

Last edited: Aug 5, 2008
2. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,073
1,774
You might want to recheck your Xc calculation.

3. ### redshaw Thread Starter Member

Jul 15, 2008
12
0
thank you for pointing that out -

i have now edited my thread with correct figures

cheers

Apr 26, 2005
4,073
1,774
5. ### redshaw Thread Starter Member

Jul 15, 2008
12
0
is it just simply

p=I^2/R

so for coil it would be 4.1W

and for capacitor 6.07W

i know the equation for pd voltage for a coil

but what is the equation for voltage pd for a capacitor??

any help much appreciated

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,073
1,774
what is pd? power disipation? Potential Drop?

I originally thought you were talking about the potential acrossed the components when you wanted the phase angle as well. That led me to recomment that page in the ebook.

Being reactive, I would tend to think the power would be VA (voltamperes) and not watts unless you factor in the power factor.

Your stated power formula is incorrect.

Is this from an instructor led course or are you doing self-study?

7. ### redshaw Thread Starter Member

Jul 15, 2008
12
0
i am doing an hnc self study course i recieve assignments through post with a list of books to read - however i am serving in the royal air force and access to books when i am away is limited.

reading through previous questions when they talk about pd it refers to voltage

i have done these calculations

Vc=IXc=13.9*31.83
Vc=442.437V

Zcoil=square root of R^2+Xl^2
=47.79

Vcoil=IZcoil=13.9*47.79
=664.34V

I am terrible at electrics as i am a mechancial engineer

8. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,073
1,774
You have the potential differences (voltage drops) slightly incorrect, you need to watch your rounding more carefully.

Attempt to keep your decimal places consistent throughout. The coil/resistor impedance is 47.798 ... which you could round up to 47.80 if your going for two decimal places. In the scheme of reality, when you consider your measuring instrument tolerances, 47.79 is within those tolerances. In academia, your answer could be incorrect.

Last edited: Aug 6, 2008