Question

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

Solution

## 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 $$i R$$ ). The resistor network can be reduced to an equivalence of$$R_{\mathrm{eq}}=\dfrac{(2 R)(R)}{2 R+R}+\dfrac{(R)(R)}{R+R}=\dfrac{7}{6} R$$which means that we can determine the current through the battery (and also through each of the bottom resistors):$$2 i=\dfrac{\varepsilon}{R_{\mathrm{eq}}} \Rightarrow i=\dfrac{\varepsilon}{2 R_{\mathrm{q}}}=\dfrac{\varepsilon}{2(7 R / 6)}=\dfrac{3 \varepsilon}{7 R}$$By the loop rule (going around the left loop, which includes the battery, resistor $$2 R$$, and one of the bottom resistors), we have$$\varepsilon-i_{2 R}(2 R)-i R=0 \Rightarrow i_{2 R}=\dfrac{\varepsilon-i R}{2 R}$$Substituting $$i=3 \varepsilon / 7 R,$$ this gives $$i_{2 R}=2 \varepsilon / 7 R$$. The difference between $$i_{2 R}$$ and $$i$$ is the current through the ammeter. Thus,$$i_{\text {ammeter }}=i-i_{2 R}=\dfrac{3 \varepsilon}{7 R}-\dfrac{2 \varepsilon}{7 R}=\dfrac{\varepsilon}{7 R} \Rightarrow \dfrac{i_{\text {ammeter}}}{\varepsilon / R}=\dfrac{1}{7}=0.143$$PhysicsNCERTStandard XII

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