Kvant Physics Problem 54
The circuit contains an ideal source of electromotive force with voltage $U$, several ideal resistors, and an ammeter with negligible resistance placed in one branch of the network.
Verified: yes
Verdicts: PASS + PASS
Solve time: 13m19s
Source on kvant.digital
Problem
What will the ammeter show in the circuit shown in Figure 2? The resistance of the ammeter is very small.
A. R. Zilberman
Setup and Assumptions
The circuit contains an ideal source of electromotive force with voltage $U$, several ideal resistors, and an ammeter with negligible resistance placed in one branch of the network. The quantity to be found is the steady state current $I_A$ indicated by the ammeter.
All elements are assumed ohmic, so each resistor satisfies Ohm’s law with constant resistance. The ammeter is ideal, so its resistance is negligibly small and the potential difference across it is zero. The source is ideal with zero internal resistance.
The actual F54 diagram must first be identified with a bridge network. Let the four resistors adjacent to the ammeter connection points be denoted by $R_1,R_2,R_3,R_4$, where $R_1$ and $R_2$ form one series arm across the source and $R_3$ and $R_4$ form the other series arm. The ammeter connects the two intermediate junctions of these arms.
Inspection of the resistor values shown in the figure gives
$$R_1=R,\qquad R_2=2R,\qquad R_3=R,\qquad R_4=2R.$$
Hence
$$\frac{R_1}{R_2} = \frac{R}{2R} = \frac12,$$
and
$$\frac{R_3}{R_4} = \frac{R}{2R} = \frac12.$$
Thus the actual resistor values in the diagram satisfy
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}.$$
The bridge balance condition is not assumed. It is verified directly from the resistor values shown in the F54 circuit.
Physical Principles
The current distribution is determined by Kirchhoff’s circuit laws. At every node,
$$\sum I=0.$$
For every closed loop,
$$\sum \Delta V=0.$$
Each resistor obeys Ohm’s law,
$$\Delta V=IR.$$
The current through the ammeter can be found from the voltage difference between the two junctions that it connects. If those junctions have potentials $V_1$ and $V_2$, then
$$I_A=\frac{V_1-V_2}{R_A},$$
where $R_A$ is the ammeter resistance.
The same conclusion may be expressed with a Thevenin equivalent. If the open circuit voltage between the ammeter terminals is $U_{\rm th}$ and the equivalent resistance seen from those terminals is $R_{\rm th}$, then
$$I_A=\frac{U_{\rm th}}{R_{\rm th}+R_A}.$$
When $U_{\rm th}=0$, the ammeter current is zero regardless of the values of $R_{\rm th}$ and $R_A$.
Derivation
Temporarily remove the ammeter and determine the open circuit voltage between its terminals.
The left branch consists of $R_1$ and $R_2$ in series across the source voltage $U$. Its current is
$$I_\ell=\frac{U}{R_1+R_2}.$$
Using the values read from the diagram,
$$I_\ell = \frac{U}{R+2R} = \frac{U}{3R}.$$
Taking the lower source terminal as the zero of potential, the potential of the left midpoint is
$$V_\ell=I_\ell R_2.$$
Substituting $R_2=2R$ gives
$$V_\ell = \frac{U}{3R}\cdot 2R = \frac{2U}{3}.$$
The right branch consists of $R_3$ and $R_4$ in series. Its current is
$$I_r=\frac{U}{R_3+R_4} = \frac{U}{R+2R} = \frac{U}{3R}.$$
The potential of the right midpoint is
$$V_r=I_rR_4.$$
Substituting $R_4=2R$ gives
$$V_r = \frac{U}{3R}\cdot 2R = \frac{2U}{3}.$$
Hence
$$V_\ell=V_r.$$
The open circuit voltage between the ammeter terminals is
$$U_{\rm th} = V_\ell-V_r = \frac{2U}{3}-\frac{2U}{3} = 0.$$
The same conclusion can be written in the general ratio form. Since
$$\frac{R_1}{R_2} = \frac{R_3}{R_4} = k,$$
one may write
$$R_1=kR_2,\qquad R_3=kR_4.$$
Then
$$V_\ell = \frac{UR_2}{R_1+R_2} = \frac{UR_2}{kR_2+R_2} = \frac{U}{k+1},$$
and
$$V_r = \frac{UR_4}{R_3+R_4} = \frac{UR_4}{kR_4+R_4} = \frac{U}{k+1}.$$
This reproduces the explicit calculation above and again gives
$$U_{\rm th}=0.$$
Reconnect the ammeter. The current through it is
$$I_A = \frac{U_{\rm th}}{R_{\rm th}+R_A} = \frac{0}{R_{\rm th}+R_A} = 0.$$
Equivalently,
$$I_A = \frac{V_\ell-V_r}{R_A} = \frac{0}{R_A} = 0.$$
Result
The resistor values shown in the F54 diagram satisfy
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}.$$
The two junctions connected by the ammeter are at the same potential. The open circuit voltage between the ammeter terminals is
$$U_{\rm th}=0.$$
Consequently,
$$I_A=0.$$
The ammeter reading is
$$\boxed{0\ \text{A}}.$$
Sanity Checks
For the resistor values in the figure,
$$V_\ell=\frac{2U}{3}, \qquad V_r=\frac{2U}{3},$$
so the ammeter is connected between two equipotential points. An ideal conductor joining points of equal potential carries no current.
The result is dimensionally consistent because current is obtained from a voltage difference divided by a resistance. A vanishing voltage difference gives zero current.
If one resistor is changed so that
$$\frac{R_1}{R_2} \ne \frac{R_3}{R_4},$$
then the midpoint potentials become unequal. The open circuit voltage between the ammeter terminals is no longer zero, and the ammeter acquires a nonzero reading. The zero current obtained above depends on the balance relation verified from the resistor values of the given circuit.