Solve Electric Circuits Homework: RT, Vdrops, I, P, VTerm

[Inertialforce]

Homework Statement

Find all of the following for the following circuit.

Resistors R1 and R2 and R6 and R7 are in series. While resistors R3, R4, and R5 are parallel.

Given information:
ε = 15.0V
ri = 0.25Ω
All other R's = 2.0Ω


1)RT or R[total] (this includes the ri)
2)Vdrops across R6
3)I (current) flow through R4
4)P dissipated by R5
5)VTerm or V[Terminal]

Homework Equations

The Attempt at a Solution

For #1 where I am supposed to find the RT or R[total], do I just add together all the R and the ri? And if so would an answer of Rt = 14.25 be correct? Because there are seven resistors each with a value of 2.0Ω so I went (7)(2.0Ω ) + (0.25) = 14.25Ω (the 0.25 being the ri).

I am kind of at a loss as to how to solve this problem because there are so many things we are asked to find. Could someone suggest to me an order I should do these questions into make things easier?

 

[Vagrant]

First of all you need to understand how resistances are combined in series and parallel. The following link has a good summary.
http://www.antonine-education.co.uk/Physics_AS/Module_3/Topic_3/topic_3__series_and_parallel_cir.htm"

This explains how to calculate total resistance in a combined circuit.
http://www.electronics-tutorials.ws/resistor/res_5.html"

 

[Inertialforce]

Could someone help me with #2 (Finding voltage drop across R6) or give me some pointers on how to find it, I am having a bit of trouble with it.

 

[LowlyPion]

I don't see your diagram.

But in general once you have derived the R_total and you determine the current (V = IR) then you are in a position to answer what the voltage across an individual resistor may be.

If they are in series then the current through one is the same as through the other. If they are in parallel then the current will be divided depending on their individual values.

 

[Inertialforce]

Oh, okay thanks. I think I understand now.

But for #4 (power dissipated by R5) can I just go:

P = I^2R
P5 = I^2R5
P5 = (15.0/5.6)^2 x (2.0) ?

The "I" here being Itotal.

Or would there be more to it because R5 is in series with R6 but parallel to R4. And R4 itself is parallel to R3 but in series with R2. While the rest of the resistors (R1 and R7 together with R3) form a normal circuit where everything is in series. So basically its a combination circuit.

 

[LowlyPion]

Oh, okay thanks. I think I understand now.

But for #4 (power dissipated by R5) can I just go:

P = I^2R
P5 = I^2R5
P5 = (15.0/5.6)^2 x (2.0) ?

The "I" here being Itotal.

Or would there be more to it because R5 is in series with R6 but parallel to R4. And R4 itself is parallel to R3 but in series with R2. While the rest of the resistors (R1 and R7 together with R3) form a normal circuit where everything is in series. So basically its a combination circuit.

The power dissipated by a resistor will be the product of the V*I across that resistor.

The current you use however is only the current through that resistor and of course you only use the voltage across that resistor.

Now you always have 2 choices of how to calculate the V*I

P = I²*R

or

P = V²/R