a) [tex]400 \Omega[/tex]
b) 0.43 V
c) 0.44 %
Explanation:
a)
For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is
[tex]V=E-Ir[/tex] (1)
where
V is the terminal voltage
E is the emf of the battery
I is the current
r is the internal resistance
In this problem, we have two situations:
1) when [tex]R_1=550 \Omega[/tex], [tex]V_1=0.25 V[/tex]
Using Ohm's Law, the current is:
[tex]I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A[/tex]
2) when [tex]R_2=1000 \Omega[/tex], [tex]V_2=0.31 V[/tex]
Using Ohm's Law, the current is:
[tex]I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A[/tex]
Now we can rewrite eq.(1) in two forms:
[tex]V_1 = E-I_1 r[/tex]
[tex]V_2=E-I_2 r[/tex]
And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:
[tex]V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega[/tex]
b)
To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)
[tex]V=E-Ir[/tex]
where
V is the terminal voltage
E is the emf of the battery
I is the current
r is the internal resistance
Using the first set of data,
[tex]V=0.25 V[/tex] is the voltage
[tex]I=4.5\cdot 10^{-4}A[/tex] is the current
[tex]r=400\Omega[/tex] is the internal resistance
Solving for E,
[tex]E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V[/tex]
c)
In this part, we are told that the area of the cell is
[tex]A=4.0 cm^2[/tex]
While the intensity of incoming radiation (the energy received per unit area) is
[tex]Int.=5.5 mW/cm^2[/tex]
This means that the power of the incoming radiation is:
[tex]P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W[/tex]
This is the power in input to the resistor.
The power in output to the resistor can be found by using
[tex]P'=I^2R[/tex]
where:
[tex]R=1000 \Omega[/tex] is the resistance of the resistor
[tex]I=3.1\cdot 10^{-4} A[/tex] is the current on the resistor (found in part A)
Susbtituting,
[tex]P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W[/tex]
Therefore, the efficiency of the cell in converting light energy to thermal energy is:
[tex]\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%[/tex]
