How to calculate detector voltage from resistances?

[PhizKid]

Homework Statement

The resistances:
Ra = 10 ohm
Rb = 10 ohm
Rc = 10 ohm
Rd = 8 ohm

Battery = 10 volts

Voltage across detector = ?

Homework Equations

V = IR
(series) R = R1 + R2 + ...
(parallel) 1 / R = (1/R1) + (1/R2) + ...

The Attempt at a Solution

So to get the voltage, I need to find the total resistance and current. I tried to simplify the problem by calculating Ra and Rb in parallel, Rc and Rd in parallel, then adding up these resistances in series. So I get the total resistance to be:

Ra + Rb = $\frac{1}{\frac{1}{10} + \frac{1}{10}}= 5$ ohm
Rc + Rd = $\frac{1}{\frac{1}{8} + \frac{1}{10}} = \frac{40}{9}$ ohm

Ra + Rb + Rc + Rd = $5 + \frac{40}{9} = \frac{85}{9}$ ohm

So to get the current, I use V/R = I
Since battery is 10V:

$\frac{10}{\frac{85}{9}} = \frac{18}{17}A$

So now I need to find the voltage through this resistance, but V = RI doesn't work because I just get back the total voltage from the battery. Not sure what else to do here

 

[mfb]

Ra and Rb (and Rc and Rd) are not in parallel. The (ideal) voltage detector does not conduct current.

 

[PhizKid]

Is Ra and Rc (Rb and Rd) in series?

 

[mfb]

Right.

 

[PhizKid]

OK, so if I add up the series first, can I then calculate the remaining 2 resistances in parallel?

 

[mfb]

Sure.

 

[willem2]

OK, so if I add up the series first, can I then calculate the remaining 2 resistances in parallel?

You can. but it won't help. You have to calculate the voltage at the point between Ra and Rc, and at the point between Rb and Rd.

Ra and Rc form a voltage divider. You should know a formula for that . If yoiu don't, caculate the current throug Ra and Rc, and then the voltage drop across Rc.

Same thing for Rb and Rd

 

[PhizKid]

I'm not sure I understand the question. If there's no current through the detector, how can there be a voltage?

 

[CWatters]

No current flows through an ideal voltmeter but it can still measure voltage. An ideal meter has very large resistance (aka infinite) so negligible current causes a voltage drop - if you insist on thinking about it that way.

I prefer to look at your diagram like this... You now have two Potential Divider chains. What is vd ?