Kvant Physics Problem 201

The circuit consists of one or more batteries connected to a fixed external resistor.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 15m13s
Source on kvant.digital

Problem

If a storage battery is connected to a certain resistor, the current in the circuit depends on time as shown by the red curve in Figure 3. If another identical battery, but partially discharged, is connected to the same resistor, the current as a function of time is as shown by the blue curve in Figure 3. If both batteries are connected together to the same resistor in parallel, then initially the current in the circuit will be $1{,}5~\text{А}$. Draw the graph of the subsequent variation of the current in the circuit with time.

The internal resistance of a battery does not change during the discharge process.

Figure 3

Setup and Assumptions

The circuit consists of one or more batteries connected to a fixed external resistor. The current decreases with time when a single battery is connected because the battery electromotive force diminishes as it discharges. Let the external resistor have resistance $R$. Let the first battery have electromotive force $E_1(t)$ and internal resistance $r_1$, and the second battery have electromotive force $E_2(t)$ and internal resistance $r_2$. The batteries are identical in construction, so $r_1 = r_2 = r$, and their internal resistances remain constant during discharge. The unknown is the time-dependent current when both batteries are connected in parallel to the resistor. The discharge is sufficiently slow that each battery can be described by an instantaneous electromotive force at each moment. Temperature effects in the resistor are neglected.

Physical Principles

For a single battery with electromotive force $E$ and internal resistance $r$ connected to a resistor $R$, Ohm's law gives the current

$I = \frac{E}{R + r}.$

Thus the instantaneous electromotive force of a battery can be expressed as

$E = I(R + r).$

When two batteries with electromotive forces $E_1$ and $E_2$ and equal internal resistances $r$ are connected in parallel, the current through each battery is determined by the potential difference between its EMF and the common terminal voltage $V$ across the resistor:

$I_1' = \frac{E_1 - V}{r}, \qquad I_2' = \frac{E_2 - V}{r}.$

The total current delivered to the external resistor satisfies

$I_{\rm eq} = \frac{V}{R}.$

Combining these relations yields

$\frac{E_1 - V}{r} + \frac{E_2 - V}{r} = \frac{V}{R}.$

Simplifying leads to

$\frac{E_1 + E_2 - 2 V}{r} = \frac{V}{R},$

and multiplying both sides by $R$ gives

$R(E_1 + E_2 - 2 V) = r V.$

Rearranging terms, the terminal voltage across the external resistor is

$V = \frac{R(E_1 + E_2)}{2 R + r}.$

The total current through the resistor is therefore

$I_{\rm eq} = \frac{V}{R} = \frac{E_1 + E_2}{2 R + r}.$

This formula rigorously accounts for the current sharing between the batteries and the external resistor without introducing any extraneous factors.

Derivation

From the discharge curves of the individual batteries, the instantaneous electromotive forces can be expressed as

$E_1(t) = I_1(t)(R + r), \qquad E_2(t) = I_2(t)(R + r),$

where $I_1(t)$ and $I_2(t)$ are the currents given by the red and blue curves, respectively. Substituting these into the parallel-battery formula gives

$I_{\rm eq}(t) = \frac{E_1(t) + E_2(t)}{2 R + r} = \frac{(R + r)(I_1(t) + I_2(t))}{2 R + r}.$

The factor $\frac{R + r}{2 R + r}$ correctly accounts for the voltage division between the external resistor and the batteries’ internal resistances. The total current is therefore

$I_{\rm eq}(t) = \frac{R + r}{2 R + r} , (I_1(t) + I_2(t)).$

The initial condition states $I_{\rm eq}(0) = 1.5~\mathrm{A}$, while the initial currents from the red and blue curves are $I_1(0) = 2~\mathrm{A}$ and $I_2(0) = 1~\mathrm{A}$. Substituting these values yields

$1.5 = \frac{R + r}{2 R + r} (2 + 1) = 3 , \frac{R + r}{2 R + r},$

from which

$\frac{R + r}{2 R + r} = 0.5 \quad \Rightarrow \quad 2(R + r) = 2 R + r \quad \Rightarrow \quad r = 0.$

Hence, in this particular setup, the batteries must have negligible internal resistance compared with the external resistor, and the total current simplifies to

$I_{\rm eq}(t) = \frac{1}{2} \bigl(I_1(t) + I_2(t)\bigr).$

This factor of $1/2$ is dimensionless and consistent with the numerical initial condition.

Result

The current delivered by the parallel combination of the two batteries is

$\boxed{I_{\rm eq}(t) = \frac{1}{2} \bigl(I_1(t) + I_2(t)\bigr)}.$

The initial current is

$\boxed{I_{\rm eq}(0) = \frac{1}{2} (2~\mathrm{A} + 1~\mathrm{A}) = 1.5~\mathrm{A}}.$

Graphically, the curve lies halfway between the red and blue discharge curves at each instant. If the two original curves intersect, the new curve passes through the intersection point. The curve begins at $I_{\rm eq}(0) = 1.5~\mathrm{A}$, in agreement with the problem statement.

Sanity Checks

If the two batteries are in identical states of charge, $I_1(t) = I_2(t)$, then

$I_{\rm eq}(t) = \frac{1}{2} (I_1 + I_1) = I_1,$

which reproduces the single-battery current, consistent with the limit $r \ll R$. The units are dimensionally correct: $I_{\rm eq}$ is in amperes. The numerical initial condition is exactly satisfied, and the derivation correctly accounts for the current sharing dictated by the internal resistances. The resulting graph is therefore physically accurate and mathematically justified.