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I am confused about the following question concerning current and drift velocity:

Two wires of the same metal have different cross sectional areas. When they are connected to a low voltage battery, the drift velocity of the electrons inside the wires is

a) slower in the thicker wire when the wires are connected in series to the battery.

b) slower in the thinner wire when the wires are connected in series to the battery.

c) slower in the thicker wire when the wires are connected in parallel to the battery.

d) slower in the thinner wire when the wires are connected in parallel to the battery.

This appears to be a contradiction. The drift velocity is slower in a thicker wire by $I=nqvA$, but the resistance is inversely proportional to the area. In short, should I take $I$ or $V$ constant in this problem?

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    $\begingroup$ Hi @David, could you please tell us your attempt at a solution and where you are having an issue? Just posting a question with no attempt will get you nowhere. $\endgroup$ – Joshuah Heath Jun 7 '17 at 16:03
  • $\begingroup$ okay I know the drift velocity is slower in thicker wire but i am having problem determining the type of connection. as I=nqvA but the resistance will also be affected as it is inversely propotional to the area .. so when Resistance decrease if area increase , I will increase I can't get how to relate all together. $\endgroup$ – David Jun 7 '17 at 16:08
  • $\begingroup$ In short What I should make constant in my calculation I or V @JoshuahHeath $\endgroup$ – David Jun 7 '17 at 16:16
  • $\begingroup$ so Dr.@JoshuahHeath what do you think is the solution and on what basis? $\endgroup$ – David Jun 7 '17 at 16:24
  • $\begingroup$ One moment, typing it up. Also, I edited your question to reflect your confusion. $\endgroup$ – Joshuah Heath Jun 7 '17 at 16:25
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Let's split this question into manageable parts.

Let $R_1$ be the resistance of the thin wire and $R_2$ be the resistance of the thick wire. The resistance is given by

\begin{equation} R=\frac{\rho L}{A}\end{equation}

Here, $L$ is the length of the wire, $A$ is the cross-sectional area, and $\rho$ is the resistivity of the material. The wires are of the same material and I will assume they are of the same length. Hence, because $R_1$ is the thinner wire, it has a smaller area and thus $R_1$ is larger than the resistance $R_2$ Thus, the first conclusion we can make is that $R_1>R_2$.

We move onto the drift velocity, which is given by \begin{equation} v=\frac{I}{nqA} \end{equation}

where $n$ is the charge carrier density, $q$ is the charge of the electron, $A$ is the area of the wire, and $I$ is the current. Now, knowing that $V=IR$ (where $V$ is the voltage from the battery across the wire), I can rewrite the above in the form

\begin{equation} v=\frac{V}{nqAR} \end{equation}

Now, plugging in our previous equation for resistance, we find that

\begin{equation} v=\frac{V}{nq\rho L} \end{equation}

Thus, the drift velocity is directly proportional to the voltage across the wire. Now, all we have to do is find the potential difference across the thin and thick wire when they are connected in parallel and series. Because this is a simple circuits exercise, I leave the rest for you to do.

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  • $\begingroup$ so @Joshuah Heath the answer is A ? $\endgroup$ – David Jun 7 '17 at 17:03
  • $\begingroup$ @David Try solving with what I've given you. I want to point you in the right direction and fill in your gaps in knowledge. If I just gave you the answer, you wouldn't learn anything. $\endgroup$ – Joshuah Heath Jun 7 '17 at 17:05
  • $\begingroup$ Don't worry I did, I am just checking anyways Thanks for your help Dr. $\endgroup$ – David Jun 7 '17 at 17:06
  • $\begingroup$ Don't worry I just did, I was just Checking Anyways Thanks for your help. $\endgroup$ – David Jun 7 '17 at 17:09

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