Help me solve a question about coils resistance

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keni4eva

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A coil of 12Ω resistance is in parallel with a 20Ω resistance. This combination is connected in series with a third coil of 8Ω resistance. if the whole circuit is connected accross a battery having an emf of 30volts and internal resistance of 2Ω. Caculate a) the terminal voltage of the battery and b) the power in the 12Ω coil.
 

Re: PLS HELP

:| (...scratch my head here)

Very eaaaassssy!!!


for a) V(bat) = 30V it stays put! (ask Sir Kirchhoff if in doubt)

for b)

Calculate the whole current in the circuit

V(bat) = Vt = 30V

Rt = ((12Ω × 20Ω)/(12Ω + 20Ω)) + 8Ω
Rt = 15.5Ω

It = Vt/Rt = 30V/15.5Ω
It = 1.9354A

V(@8Ω) = It×R(@8Ω) = 1.9354A(8Ω)
V(@8Ω) = 15.4838V

V(@12Ω) = Vt - V(@8Ω) = 30V - 15.4838V
V(@12Ω) = 14.5161V

P = IV

but I = V/R

P = V²/R = (V@12Ω)²/(12Ω)
P = (14.5161V)²/12Ω
P = 17.5598W

Added after 4 minutes:

I suggest you to play with Ohm's Law and Kirchhoff's laws

You should know that by now, I am only helping you but please make youe study serious (but don't take too serious like people in ICUs, hey, studying is fun)
 

Re: PLS HELP

am not sure if your a is correct
because i know terminal voltage to be emf minus current mulitply by the internal resistor. emf-current*internal resistor.
 

Re: PLS HELP

That has been given in the question to be (connected accross a battery having an emf of 30volts and internal resistance of 2Ω.)
 

Re: PLS HELP

So just put 2Ω in series with every thing and you can find the answer.

Rtot = 15.5 + 2 = 17.5
Itot = 30/17.5 = 1.71A

P(12Ω) = =V*I= (I*R)*I= I(12Ω)²*12Ω

I(12Ω) = Itot(20/(12+20))= 1.71(20/32)= 1.07A
P(12Ω) = 1.07² * 12 = 13.77W

Hope this helps
 

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