Kvant Physics Problem 66

Verified: yes
Verdicts: PASS + PASS
Solve time: 8m51s
Source on kvant.digital

Problem

F66. An ammeter and a voltmeter are connected in series to a battery with an emf of 9 volts and an unknown internal resistance (Fig. 1). The resistances of the instruments are unknown. If a resistor (whose value is also unknown) is connected in parallel with the voltmeter, the ammeter reading doubles while the voltmeter reading is reduced by a factor of two. What is the voltmeter reading after the resistor is connected?

A. R. Zilberman

Fig. 1. A battery with an emf of 9 V and an unknown internal resistance is connected in series with an ammeter $A$ and a voltmeter $V$; an unknown resistor is connected in parallel with the voltmeter.

Setup and Assumptions

The circuit consists of a battery with electromotive force

$$\mathcal E = 9,\text{V},$$

internal resistance $r$, an ammeter of resistance $R_A$, and a voltmeter of resistance $R_V$. The ammeter and voltmeter are connected in series.

Initially, the current in the circuit is $I_1$, and the voltmeter reading is $U_1$.

After an additional resistor of resistance $R$ is connected in parallel with the voltmeter, the equivalent resistance of that branch becomes

$$R_p=\frac{R_VR}{R_V+R}.$$

The current becomes $I_2$, and the voltmeter reading becomes $U_2$.

The problem states that

$$I_2=2I_1,$$

and

$$U_2=\frac{U_1}{2}.$$

The quantity to be found is the voltmeter reading $U_2$ after the resistor is connected.

The instruments are treated as ordinary resistors with constant resistances. The battery emf and internal resistance remain unchanged. Steady direct current conditions are assumed.

Physical Principles

The solution uses Ohm's law and Kirchhoff's loop rule.

For a circuit carrying current $I$, the terminal voltage across a resistance $R$ is

$$U=IR.$$

For a complete circuit containing a source of emf $\mathcal E$ and total resistance $R_{\text{tot}}$,

$$I=\frac{\mathcal E}{R_{\text{tot}}}.$$

The voltmeter reading equals the potential difference across the branch in which the voltmeter is connected.

Derivation

Before the resistor is added, the ammeter resistance, voltmeter resistance, and battery internal resistance are all in series. Hence

$$I_1=\frac{\mathcal E}{r+R_A+R_V}.$$

The voltmeter measures the voltage across its own resistance:

$$U_1=I_1R_V.$$

After the resistor is connected in parallel with the voltmeter, the equivalent resistance of that branch is $R_p$. The current becomes

$$I_2=\frac{\mathcal E}{r+R_A+R_p}.$$

Since the ammeter reading doubles,

$$I_2=2I_1.$$

Substituting the expressions for the currents,

$$\frac{\mathcal E}{r+R_A+R_p} = 2\frac{\mathcal E}{r+R_A+R_V}.$$

After canceling $\mathcal E$,

$$r+R_A+R_V = 2(r+R_A+R_p).$$

Rearranging,

$$R_V=(r+R_A)+2R_p.$$

Hence

$$r+R_A=R_V-2R_p.$$

The voltmeter reading after the resistor is added is

$$U_2=I_2R_p.$$

Using the condition $U_2=U_1/2$,

$$I_2R_p=\frac12 I_1R_V.$$

Since $I_2=2I_1$,

$$2I_1R_p=\frac12 I_1R_V.$$

Cancelling $I_1$ gives

$$2R_p=\frac12 R_V,$$

so

$$R_p=\frac{R_V}{4}.$$

Substituting this relation into

$$r+R_A=R_V-2R_p,$$

yields

$$r+R_A = R_V-\frac{R_V}{2} = \frac{R_V}{2}.$$

The initial current is therefore

$$I_1 = \frac{\mathcal E} {r+R_A+R_V} = \frac{\mathcal E} {\frac{R_V}{2}+R_V} = \frac{\mathcal E}{\frac32R_V} = \frac{2\mathcal E}{3R_V}.$$

The initial voltmeter reading is

$$U_1=I_1R_V = \frac{2\mathcal E}{3}.$$

Consequently,

$$U_2=\frac{U_1}{2} = \frac{\mathcal E}{3}.$$

Result

The symbolic result is

$$U_2=\frac{\mathcal E}{3}.$$

Substituting the given emf,

$$U_2=\frac{9,\text{V}}{3}=3,\text{V}.$$

$$\boxed{U_2=3,\text{V}}$$

Sanity Checks

The result has the correct dimension of voltage, since it is obtained by multiplying a dimensionless factor by the emf $\mathcal E$.

Using $R_p=R_V/4$, the parallel branch has a smaller resistance than the original voltmeter alone, so the total circuit resistance decreases. The current should increase, consistent with the statement that the ammeter reading doubles.

From

$$U_1=\frac{2}{3}\mathcal E=6,\text{V},$$

the final reading becomes

$$U_2=3,\text{V},$$

which is exactly one half of the initial reading as required.

The most error-prone step is combining the two experimental conditions,

$$I_2=2I_1 \quad\text{and}\quad U_2=\frac{U_1}{2},$$

to obtain

$$R_p=\frac{R_V}{4}.$$

Replacing $I_2$ by $I_1/2$ instead of $2I_1$, or mishandling the factor $1/2$ in the voltage condition, would lead to an incorrect resistance ratio and an incorrect final voltage.