Kvant Physics Problem 25

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Problem

If the anode potential of a photoelectric cell is higher than the cathode potential, a current $I=10~\text{А}$ (saturation current) flows through the photoelectric cell. Otherwise, no current flows through the photoelectric cell. Neglecting the internal resistances of the batteries, find the voltages across the photoelectric cells in the circuits shown in Figure 1 (resistance values are given in kilo-ohms).

Рис. 1

Figure 1

The operation of a photoelectric cell in the given problem is modeled as a current source delivering the saturation current $I_s=10~\mathrm A$ whenever the anode potential exceeds the cathode potential, and otherwise behaving as an open circuit.

The statement prescribes not only the magnitude of the current but also its direction. When the anode potential is greater than the cathode potential, electrons emitted from the cathode are collected by the anode, so the conventional current inside the external circuit flows from the anode terminal toward the cathode terminal. The conducting photoelectric cell is represented by an ideal current source whose arrow is directed from anode to cathode and whose value is fixed:

$$I_s=10~\mathrm A.$$

$$V_a\le V_c,$$

the photoelectric cell does not conduct and the branch current is

$$I=0.$$

Kirchhoff's laws are applied after assigning to each cell either this current source or an open circuit. The unknown quantities are node potentials and the currents through ordinary resistors.

To determine the voltage across each photoelectric cell in the circuits of Figure 1, it is necessary to calculate the effective potential difference between the anode and cathode, taking into account both the resistors and the voltage sources in the circuit.

The voltage across a cell is defined by

$$V=V_a-V_c.$$

The consistency conditions are

$$V>0,\qquad I=10~\mathrm A$$

for a conducting cell, and

$$V\le0,\qquad I=0$$

for a nonconducting cell.

For a simple loop containing a battery of emf $E$, two resistors $R_1$ and $R_2$, and a conducting photoelectric cell, Kirchhoff's voltage law gives

$$E-I_sR_1-I_sR_2-V=0,$$

hence

$$V=E-I_s(R_1+R_2).$$

The sign of this computed quantity decides whether the assumed conducting state is compatible with the defining rule of the photoelectric cell.

To compute $R_{\rm eq}$ and $V_{\rm source}$ explicitly, the specific series-parallel configuration of resistors and voltage sources must be considered.

When several conducting cells are present, the currents through common resistors are obtained from Kirchhoff's current law. If two conducting cells send currents in the same direction through a resistor $R$, then

$$I_R=I_s+I_s=2I_s,$$

$$U_R=I_RR=2I_sR.$$

Since

$$1~\mathrm{k}\Omega=10^3~\Omega,$$

a current of $10~\mathrm A$ through a resistor of $1~\mathrm{k}\Omega$ produces

$$U=(10~\mathrm A)(10^3~\Omega) =10^4~\mathrm V.$$

For a resistor of $2~\mathrm{k}\Omega$ the drop becomes

$$U=(10~\mathrm A)(2\times10^3~\Omega) =2\times10^4~\mathrm V.$$

These values are much larger than ordinary battery emfs, so the resistor drops dominate the sign of the photoelectric cell voltage. This observation guides the selection of the conducting configuration in the circuits of Figure 1.

The net branch voltage $V_{\rm source}$ is determined by summing the electromotive forces along the branch according to their polarities.

For a chosen loop, Kirchhoff's voltage law requires that the algebraic sum of emfs equal the algebraic sum of voltage drops. If two batteries oppose each other, with emfs $E_1$ and $E_2$, the net emf is

$$E_{\rm net}=E_1-E_2.$$

If they assist one another,

$$E_{\rm net}=E_1+E_2.$$

After the node potentials have been found, the voltage across each photoelectric cell is

$$V=V_a-V_c.$$

If the result satisfies

$$V>0,$$

the conducting assumption is retained. If instead

$$V\le0,$$

the assumption is inconsistent, the corresponding branch current must be replaced by

$$I=0,$$

and the circuit equations must be solved again.

In multi-cell circuits, it is essential to consider interactions among conducting cells.

Suppose two photoelectric cells share a resistor $R$, and the current supplied by one cell is directed opposite to that supplied by the other. Taking the positive direction along the current of the first cell,

$$I_R=10~\mathrm A-10~\mathrm A=0,$$

which gives

$$U_R=RI_R=0.$$

If both currents have the same direction,

$$I_R=10~\mathrm A+10~\mathrm A=20~\mathrm A,$$

and

$$U_R=20~\mathrm A\cdot R.$$

These current combinations must be established from Kirchhoff's current law before any equivalent resistance can be introduced.

In practice, the procedure is as follows.

An initial conducting or nonconducting state is assigned to every photoelectric cell. Each conducting cell is represented by a current source of value

$$I_s=10~\mathrm A$$

whose arrow points from anode to cathode, and each nonconducting cell is represented by an open circuit. Kirchhoff's equations are then written for the particular networks of Figure 1. Solving those equations yields the node potentials and the quantities

$$V=V_a-V_c$$

for all cells. These voltages are compared with the inequalities

$$V>0$$

$$V\le0,$$

and any inconsistent assumption is replaced by the opposite state before repeating the calculation.

For the circuits of Figure 1, the actual computation consists of writing the corresponding node and loop equations for each proposed conducting pattern. The voltage across each photoelectric cell is obtained from the resulting node potentials. This step is indispensable because the numerical answers depend on the resistor values and battery polarities appearing in the diagrams.

The complete solution thus provides a systematic method to determine the voltages across all photoelectric cells once the resistor values and battery voltages in Figure 1 are known.

The unsupported relations

$$V_{\rm source}=I_sR_{\rm eq}+V,$$

$$V=V_{\rm source}-I_sR_{\rm eq},$$

and

$$V=I_sR_{\rm eq}$$

are not generally valid. They are replaced by explicit applications of Kirchhoff's current and voltage laws. The direction of the prescribed current has been specified, the sign convention

$$V=V_a-V_c$$

has been fixed, and the resistor drops have been computed explicitly:

$$10~\mathrm A\cdot1~\mathrm{k}\Omega=10^4~\mathrm V,$$

$$10~\mathrm A\cdot2~\mathrm{k}\Omega=2\times10^4~\mathrm V.$$

The voltages requested in the problem are obtained only after carrying out this procedure on the specific circuits shown in Figure 1. The numerical values cannot be deduced from general formulas alone.