# Why do we subtract the voltage of the resistor from the emf voltage to find the terminal voltage?

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?

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

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