Compound circuit voltage drop along parallel component

In summary, the equivalent resistance of the circuit is 17.31 Ω. The voltage drop across the parallel arrangement in the upper branch is 30.0 V. The total voltage of the circuit is 51.93 V. The calculations for these values were found using the formula V = I * R. The voltage value found is correct, even though it may seem unusual compared to previous problems given.
  • #1
bnosam
148
0

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: [itex] \frac{1}{\frac{1}{15.0 Ω} + \frac{1}{30.0 Ω}} = 10.0 Ω [/itex]


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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  • #2
What makes you question the voltage? It is in the ballpark.
 
  • #3
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.
 
  • #4
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: [itex] \frac{1}{\frac{1}{15.0 Ω} + \frac{1}{30.0 Ω}} = 10.0 Ω [/itex]

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:
  • #5
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?
 
  • #6
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!
 
  • #7
SammyS said:
Yes, that's correct!

Thanks SammyS, I appreciate your help.
 

FAQ: Compound circuit voltage drop along parallel component

What is a compound circuit?

A compound circuit is a combination of resistors and other components connected in a circuit. It can be a combination of series and parallel circuits, which allows for more complex and versatile electrical systems.

What is voltage drop?

Voltage drop refers to the decrease in voltage that occurs as current flows through a circuit. This drop is caused by the resistance of the components in the circuit, and it can affect the overall performance of the circuit.

How is voltage drop calculated in a parallel circuit?

In a parallel circuit, the voltage drop across each component is the same, as the current is divided among the components. To calculate the voltage drop across a parallel component, you can use Ohm's Law (V=IR) where V is the voltage drop, I is the current, and R is the resistance of the component.

What factors can affect voltage drop in a compound circuit?

The main factor that affects voltage drop in a compound circuit is the resistance of the components. The higher the resistance, the greater the voltage drop. Other factors that can affect voltage drop include the length and thickness of the wires, temperature, and the type of material used for the components.

How can voltage drop be reduced in a compound circuit?

Voltage drop can be reduced in a compound circuit by using components with lower resistance, using thicker wires to reduce resistance, and keeping the circuit at a lower temperature. Additionally, adding more power sources or using voltage regulators can also help to reduce voltage drop.

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