I attached an image of a question from homework. I calculated the current in step 1) and then I calculated the voltage of the resistor. But why do I need to subtract that voltage from the emf voltage in order to get the terminal voltage?

enter image description here

I am new to the Physics Stack Exchange so please let me know if I cannot ask questions like this on here before downvoting.

  • $\begingroup$ You didn't just use $V=IR$ for the volt meter? $\endgroup$ – Aaron Stevens Feb 20 at 4:42
  • $\begingroup$ That would yield a different answer of 1.49779... so why is that? is it still right? $\endgroup$ – Tom el Safadi Feb 20 at 4:45
  • $\begingroup$ I think you're making a mistake... $\left (4.99918\times10^{-4}\ \rm A\right)\left (3000\ \Omega\right)=1.49975\ \rm V$ $\endgroup$ – Aaron Stevens Feb 20 at 4:50

Using Kirchoff's rule for voltage drops around the circuit $$\epsilon-v_m-v_r=0$$ where "m" and "r" represent the voltmeter and internal resistance respectively.

So you could use the given EMF and the voltage across the resistor, given by $v_r=IR_r$, so that $$v_m=\epsilon-IR_r$$ and this is why your method works.

However, it's much easier just to do $$v_m=IR_m$$

Both ways are self-consistent because $$\epsilon=I(R_m+R_r)$$ so $$v_m=\epsilon-IR_r=I(R_m+R_r)-IR_r=IR_m$$


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