Compound circuit voltage drop along parallel component

AI Thread Summary
The discussion revolves around calculating the voltage drop in a compound circuit with parallel components. The user initially calculated the total voltage as 51.93 V using the equivalent resistance of 17.31 Ω and a current of 3.00 A, but expressed uncertainty about this result. After collapsing the series and parallel resistances, they determined the voltage drop across the upper parallel arrangement to be 30.0 V, which was confirmed by other participants. The user sought clarification on their calculations and received positive feedback on their approach. The conversation highlights the importance of verifying calculations in circuit analysis.
bnosam
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Homework Statement



http://imageshack.us/a/img197/3519/circuith.png

Equivalent Resistance of circuit = 17.31 Ω


Homework Equations



V = I * R

The Attempt at a Solution



Total voltage of the circuit, if I'm doing this correctly should be:

E = (3.00 A) * (17.31 Ω) = 51.93 V

But something doesn't seem right about this number

When we collapse the series in the top parallel circuit branch it becomes: 24.0 Ω + 6.00 Ω = 30.0 Ω

Then we collapse the whole parallel circuit into a single resistor equal to: \frac{1}{\frac{1}{15.0 Ω} + \frac{1}{30.0 Ω}} = 10.0 Ω


I'm not quite sure where to go past this and I'm questioning the voltage I'm getting, so any pointers in the right direction would be awesome, thanks :)
 
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What makes you question the voltage? It is in the ballpark.
 
lewando said:
What makes you question the voltage? It is in the ballpark.

The thing that makes me question the voltage is the fact that all the voltages given to us so far in any problems were whole numbers and usually over 100V.

And if that's the correct voltage, then I'm not quite sure where to go next.
 
bnosam said:

Homework Statement



http://imageshack.us/a/img197/3519/circuith.png

Equivalent Resistance of circuit = 17.31 Ω

Homework Equations



V = I * R

The Attempt at a Solution



Total voltage of the circuit, if I'm doing this correctly should be:

E = (3.00 A) * (17.31 Ω) = 51.93 V

But something doesn't seem right about this number

When we collapse the series in the top parallel circuit branch it becomes: 24.0 Ω + 6.00 Ω = 30.0 Ω

Then we collapse the whole parallel circuit into a single resistor equal to: \frac{1}{\frac{1}{15.0 Ω} + \frac{1}{30.0 Ω}} = 10.0 Ω

I'm not quite sure where to go past this and I'm questioning the voltage I'm getting, so any pointers in the right direction would be awesome, thanks :)
What is it you are trying to find?

The voltage value you found is correct.
 
Last edited by a moderator:
SammyS said:
What is it you are trying to find?

The voltage value you found is correct.

I type that up and I managed to forget the actual question! Haha, sorry.

The question:

What is the voltage drop across the parallel arrangement in the upper branch?

E = I * R

E = (3.00 A) * 10.0 Ω = 30.0 V?

That looks right to me. Anyone confirm for me please?
 
bnosam said:
I type that up and I managed to forget the actual question! Haha, sorry.

The question:

What is the voltage drop across the parallel arrangement in the upper branch?

E = I * R

E = (3.00 A) * 10.0 Ω = 30.0 V?

That looks right to me. Anyone confirm for me please?

Yes, that's correct!
 
SammyS said:
Yes, that's correct!

Thanks SammyS, I appreciate your help.
 
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