Parallel Circuits - Joining 2 branches together

[wolfsloth]

Homework Statement

I'm stuck at part (b)

Homework Equations

V = IR (ohm's law)

V1 = (R1 / (R1 + R2) ) * V (potential divider)

The Attempt at a Solution

I started off by thinking which direction conventional current would flow.

I concluded that the potential at B will be higher than the potential at E, so the current will flow from B to E. Thus, the current will not flow through the 3 Ω resistor.

This was the new circuit diagram I drew (where I thought the current would flow now):

However, the answers I got for PD across AB and DE were wrong.

So I think my concept of the direction of current flow is wrong.

 

[BvU]

Hello wolfsloth,

I see two pictures of the new situation: one has 4.0 $\Omega$ and the other 1 $\Omega$ on the right. You continue with the 1 $\Omega$, as if the 3 $\Omega$ is absent.

I concluded that the potential at B will be higher than the potential at E

The points B and E are joined by a wire. A wire (in most exercises) has no resistance, so the potential on one end is the same as on the other end. I e. both the 3 and the 1 'see' the same potential difference: they are effectively in parallel, just like the 2 and the 4 on the left.

 

[CWatters]

I would forget about trying to work out where currents are going for the moment. Instead try and simplify the circuit first. On the left you have a 2R and a 4R in parallel. On the right you have a 3R and 1R in parallel. Replace the 2R//4R with one equivalent resistor. Replace the 3R//1R with another equivalent resistor. Redraw the new circuit.

 

[wolfsloth]

Hello wolfsloth,

I see two pictures of the new situation: one has 4.0 $\Omega$ and the other 1 $\Omega$ on the right. You continue with the 1 $\Omega$, as if the 3 $\Omega$ is absent.
The points B and E are joined by a wire. A wire (in most exercises) has no resistance, so the potential on one end is the same as on the other end. I e. both the 3 and the 1 'see' the same potential difference: they are effectively in parallel, just like the 2 and the 4 on the left.

I would forget about trying to work out where currents are going for the moment. Instead try and simplify the circuit first. On the left you have a 2R and a 4R in parallel. On the right you have a 3R and 1R in parallel. Replace the 2R//4R with one equivalent resistor. Replace the 3R//1R with another equivalent resistor. Redraw the new circuit.

Hi everyone, thank you for the reply!

Unfortunately, seeing the two resistors on the left as connected in parallel and the two resistors on the right as connected in parallel gives me the wrong answer :(

Here are my workings:

Correct answer should be 0.8V for PD across AB and 1.6V for PD across DE

 

[wolfsloth]

Hello wolfsloth,

I see two pictures of the new situation: one has 4.0 $\Omega$ and the other 1 $\Omega$ on the right. You continue with the 1 $\Omega$, as if the 3 $\Omega$ is absent.
The points B and E are joined by a wire. A wire (in most exercises) has no resistance, so the potential on one end is the same as on the other end. I e. both the 3 and the 1 'see' the same potential difference: they are effectively in parallel, just like the 2 and the 4 on the left.

Also, if you refer to part (ii), there is a potential difference across points B and E :O

 

[BvU]

Correct answer should be 0.8V for PD across AB and 1.6V for PD across DE

Your working is correct. There must be something wrong with the book answer -- it can not be that the two answers differ.

 

[CWatters]

+1

I agree. Your working is 100% correct. There is definitely something wrong with the question and/or the answer.

With BE joined the voltage AB and DE should be the same. That's because nodes A and D are connected together and B and E are connected together.

 

[BvU]

The 0.8 V for AB and 1.6 V for DE are the values before B and E are joined by a wire.

So I suspect the line ' (b) The points B and E are now joined by a wire ' is misplaced and should have been on the next page