Question

A solar cell generates a potential difference of $0.10\;V$ when a $500\;\Omega$ resistor is connected across it, and a potential difference of $0.15\;V$ when a $1000\;\Omega$ resistor is substituted. What are the a) internal resistance b) emf of the cell c) The area of the cell is $5\;c{m^2}$, and the rate per unit area at which it receives energy from light is $2.0\;mW/c{m^2}$. What is the efficiency of the cell for converting light energy into thermal energy in the $1000\;\Omega$ external resistor?

Answer 1

The formula for internal resistance is applied. By using the pair of potential difference and resistor values given in the question, two equations of internal resistance are constructed and the value of internal resistance and the emf value is found out by solving the simultaneous equations. The efficiency is given by the ratio of the value of power of the $1000\;\Omega$ resistor to that of the power of the incident light received by the solar cell.

Formula used:
The formula for the internal resistance is as follows:
$r = R\left[ {\dfrac{{E - V}}{V}} \right]$ ----($1$)
where, $r$ is the internal resistance, $E$ is the emf, $V$ is the terminal potential difference and $R$ is the resistance.

Complete step by step answer:
The internal resistance is defined as the resistance that is offered by the electrolyte of a cell to the flow of current between its electrodes.
When $V = 0.10\;V$a resistor of resistance $R = 500\;\Omega$ is connected and
When $V = 0.15\;V$ a resistor of resistance $R = 1000\;\Omega$ is connected
Hence, two equations of internal resistance can be constructed by applying the formula from equation ($1$). The equations are as follows:
$r = 500\left[ {\dfrac{{E - 0.10}}{{0.10}}} \right]$ ----($2$)
$\Rightarrow r = 1000\left[ {\dfrac{{E - 0.15}}{{0.15}}} \right]$ ----($3$)

(b) Since there are two equations and two unknown variables the equations can be solved simultaneously to find out the values of internal resistance and emf.
From the equations ($2$) and ($3$) we get:
$r = 5000\left[ {E - 0.10} \right]$
$\Rightarrow r = \dfrac{{20000}}{3}\left[ {E - 0.15} \right]$
The equations are further simplified to get:
$r = 5000E - 500$ ----($4$)
$\Rightarrow r = 6666.7E - 1000$ -----($5$)
Subtracting equation ($4$) from ($5$) we get:
$0 = 1666.7E - 500$
By rearranging the terms we calculate the value of emf which is:
$\therefore E = 0.299\;V$
After rounding off to three significant figures we get:
$E \approx 0.30\;V$

(a) Substituting the value of emf in one of the equations we get:
$r = 5000(0.30) - 500$
$\therefore r = 1000\;\Omega$

Hence we get the value of internal resistance to be $1000\;\Omega$ and the value of emf to be $0.30\;V$.

(c) The efficiency in the process of converting light energy to thermal energy in the external resistor is determined by finding the ratio of the power generated by the $1000\;\Omega$ resistor by the intensity of the incident light which is the power that is received by the solar cell. First, we find the power in the $1000\;\Omega$ resistor. This is given by the formula:
$P = \dfrac{{{V^2}}}{R}$ where the variables denote the quantities already mentioned above.
$\Rightarrow P = \dfrac{{0.15 \times 0.15}}{{1000}}$
$P\approx 2.25 \times {10^{ - 5}}\;W$ ----($6$)
The area of the cell that is given is: $5\;c{m^2}$
The rate of absorption of the intensity of light per unit area that is given is: $2.0\;mW/c{m^2}$
This intensity is converted from square meters to square centimeters.
This will be equivalent to $2.0 \times {10^{ - 3}}\;W/c{m^2}$.
There is a formula to calculate the intensity received by the solar cell.

The light energy absorbed by the solar cell per second is $=$ area of the cell $\times$ intensity of light
$\Rightarrow {P_{received}}=5 \times 2 \times {10^{ - 3}}\;W$
$\Rightarrow {P_{received}}=10 \times {10^{ - 3}}\;W$
$\Rightarrow {P_{received}} = {10^{ - 2}}\;W$ ----($7$)
The efficiency as explained above is given by the ratio:
$efficiency = \dfrac{{{P_{1000\;\Omega }}}}{{{P_{received}}}} \times 100\%$
Hence, substituting the values obtained in equation ($6$) and ($7$) we get:
$\text{efficiency} = \dfrac{{2.25 \times {{10}^{ - 5}}}}{{{{10}^{ - 2}}}} \times 100\%$
$\therefore \text{efficiency} = 0.225\%$
The value is rounded off to three significant figures giving: $\approx 0.23\%$

Hence the efficiency of the cell for converting light energy into thermal energy in the $1000\;\Omega$ external resistor is $0.23\%$.

Additional information: In a circuit, the terminals are connected by wires through which current flows through its conventional direction that is from positive to negative terminal. But inside the electrolyte of the cell, the positive ions have a tendency to flow from higher to lower potential (and vice-versa for negative ions) against other ions and neutral atoms present in the electrolyte. Hence, the electrolyte is responsible for offering home resistance to the flow of current in the cell. This phenomenon is the concept behind internal resistance.

The work done in carrying a unit charge along a closed circuit is known as EMF while the potential drop across the terminals of the cell when there is a current drawn from it is called potential difference. When there is no current drawn from the cell then these two quantities become equal. However, in reality there is always some internal resistance and hence potential difference is lesser than the EMF of the circuit.

Note: The EMF of the circuit is not equal to the terminal potential difference of a cell in a closed circuit which is usually mistaken. The potential difference across the terminals of a cell is always lesser than the value of the emf of the cell when the circuit is closed.