A solar cell generates a potential difference of $\mathbf{\text{0.10V}}$when a$\mathbf{\text{500}}\mathit{\Omega }$ resistor is connected across it, and a potential difference of $\mathbf{\text{0.15V}}$when a $\mathbf{\text{1000}}\mathit{\Omega }$resistor is substituted.

(a) What is the internal resistance?

(b) What is the emf of the solar cell?

(c) The area of the cell is${\mathbf{\text{5.0cm}}}^{\mathbf{\text{2}}}$ , and the rate per unit area at which it receives energy from light is${\mathbf{\text{2.0mW/cm}}}^{\mathbf{\text{2}}}$ .What is the efficiency of the cell for converting light energy to thermal energy in the$\mathbf{\text{1000}}\mathit{\Omega }$ external resistor?

$$\begin{gathered} V=\text{0.10V} \\ R=500\text{Ω} \\ {V}^{\text{'}}=\text{0.15V} \\ {R}^{\text{'}}=1000\text{Ω} \\ A={\text{5.0cm}}^{\text{2}}{\text{=5×10}}^{\text{-4}}{\text{m}}^{\text{2}} \\ \frac{P_{input}}{A}=2.\text{0}\frac{\text{mW}}{{\text{cm}}^{\text{2}}}\text{=20}\frac{\text{W}}{{\text{m}}^{\text{2}}} \end{gathered}$$

Write two expressions for emf of the solar cell for both the voltages and the resistances using Ohm’s law. Equating them can give the internal resistance. Inserting the value of internal resistance in one of the equation will give emf of the cell. Then from the rate of the energy received per unit area, find the efficiency using the corresponding formula.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points

Formulae are as follow:

$I=\frac{V}{R}$

Where, I is current, V is voltage, R is resistance.

Let the internal resistance of the battery be r and emf be$\epsilon .$

When the resistance R is in the circuit, emf is

$\epsilon =V+Ir$

But, according to the Ohm’s law,

$$\begin{gathered} I=\frac{V}{R} \\ \therefore \epsilon =0.10+\frac{0.10}{500}r \\ \therefore \epsilon =0.10+2\times {10}^{-4}r\dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots 1) \end{gathered}$$

When resistance R’ is in the circuit, emf is,

$\epsilon =V\text{'}+Ir$

But, according to the Ohm’s law,

$$\begin{gathered} I=\frac{V\text{'}}{R\text{'}} \\ \therefore \epsilon =0.15+\frac{0.15}{1000}r \\ \therefore \epsilon =0.15+1.5\times {10}^{-4}r\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots 1) \end{gathered}$$

Equating 1) and 2) gives,

$r=1.0\times {10}^3\text{Ω}$

Hence,the internal resistance of the solar cell is $1.0\times {10}^3\text{Ω}$.

Inserting value of r in equation 1 yields,

$\epsilon =0.10+2\times {10}^{-4}(1.0\times {10}^3)$

$\epsilon =\text{0.30V}$

Hence, the emf of the solar cell is$\text{0.30V.}$

Efficiency of the solar cell is,

$$\begin{gathered} \eta =\frac{P_{output}}{P_{input}}=\frac{\frac{V^2}{R}}{P_{input}} \\ \eta =\frac{\frac{{\left(0.15\right)}^2}{1000}}{20\times 5\times {10}^{-4}} \\ \eta =2.25\times {10}^{-3}~0.23\% \end{gathered}$$

Hence, the efficiency of the solar cell is$0.23\%$

Therefore, the internal resistance, emf and the efficiency of the solar cell from the potential difference, the resistance across it and the power received can be determined using Ohm’s law and the corresponding formulae.