Question

In Figure , $R_1=2.00R$, the ammeter resistance is zero, and the battery is ideal. What multiple of $E$/R gives the current in the ammeter?

Answer 1

Note that there is no voltage drop across the ammeter. Thus, the currents in the bottom resistors are the same, which we call $i$ (so the current through the battery is $2i$ and the voltage drop across each of the bottom resistors is $iR$ ). The resistor network can be reduced to an equivalence of

$R_{eq}=\frac{\left(2R\right)\left(R\right)}{2R+R}+\frac{\left(R\right)\left(R\right)}{R+R}=\frac{7}{6}R$

which means that we can determine the current through the battery (and also through each of the bottom resistors):

$2i=\frac{\epsilon }{R_{eq}}\Rightarrow i=\frac{\epsilon }{2R_q}=\frac{\epsilon }{2\left(7R/6\right)}=\frac{3\epsilon }{7R}$

By the loop rule (going around the left loop, which includes the battery, resistor $2R$, and one of the bottom resistors), we have

$\epsilon -i_{2R}\left(2R\right)-iR=0\Rightarrow i_{2R}=\frac{\epsilon -iR}{2R}$

Substituting $i=3\epsilon /7R,$ this gives $i_{2R}=2\epsilon /7R$. The difference between $i_{2R}$ and $i$ is the current through the ammeter. Thus,

$i_{\text{ammeter}}=i-i_{2R}=\frac{3\epsilon }{7R}-\frac{2\epsilon }{7R}=\frac{\epsilon }{7R}\Rightarrow \frac{i_{\text{ammeter}}}{\epsilon /R}=\frac{1}{7}=0.143$