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I have the following circuit:

The voltmeter \$V_1\$ measures the voltage drop across the \$9.00 \ \Omega\$ resistor and has a reading of \$4.50 \ \text{V}\$. enter image description here

The reading on ammeter \$A_1\$ is \$I_1 = 0.50 \ \text{A}\$, and the reading on ammeter \$A_2\$ is \$I_2 = 0.38 \ \text{A}\$.

I am now trying to determine the unknown resistance \$R\$. It seems to me that we have two resistors in parallel here, so we can use the formula \$R_{\text{total}} = \dfrac{R_1 R_2}{ R_1 + R_2 }\$. So we have that \$R_{\text{total}} = \dfrac{R \times 4.0 \ \Omega}{ R + 4.0 \ \Omega }\$. We can now incorporate this into Ohm's law, where \$I = 0.38 \ \text{A}\$ and \$V = 6.0 \ \text{V} - 4.5 \ \text{V} = 1.5 \ \text{V}\$, so that we get \$1.5 = 0.38 \left( \dfrac{R \times 4.0 \ \Omega}{ R + 4.0 \ \Omega } \right) \ \Rightarrow R = 300 \ \Omega\$. But I am told that the solution is actually \$ I = 0.5 - 0.375 = 0.125 \ \text{A} \$ (where I round \$0.375\$ to \$0.38\$) and \$R = \dfrac{1.5}{0.125} = 12 \ \Omega\$. How is my reasoning incorrect, and what is the correct reasoning that leads to \$ I = 0.5 - 0.375 = 0.125 \ \text{A} \$ and \$R = \dfrac{1.5}{0.125} = 12 \ \Omega\$?

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It looks like you do not see what path the current is taking in this circuit. This may be handy enter image description here

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  • \$\begingroup\$ But aren't we suppose to consider conventional current (current travelling clockwise)? \$\endgroup\$ May 12, 2022 at 12:58
  • \$\begingroup\$ @ThePointer No. Conventional current flows from higher (more positive) to lower potential. \$\endgroup\$
    – devnull
    May 12, 2022 at 13:03
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    \$\begingroup\$ @devnull Oops, yes, from the positive terminal of the battery to the negative terminal of the battery. \$\endgroup\$ May 12, 2022 at 13:05
  • \$\begingroup\$ @ThePointer While you can do it the other way, you have to use negative signs everywhere, and it kinda turns into a mess. It's much easier to do the math in the positive direction. \$\endgroup\$
    – Aaron
    May 12, 2022 at 13:52

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