a) ![400 \Omega](https://tex.z-dn.net/?f=400%20%5COmega)
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
(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
, ![V_1=0.25 V](https://tex.z-dn.net/?f=V_1%3D0.25%20V)
Using Ohm's Law, the current is:
![I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A](https://tex.z-dn.net/?f=I_1%3D%5Cfrac%7BV_1%7D%7BR_1%7D%3D%5Cfrac%7B0.25%7D%7B550%7D%3D4.5%5Ccdot%2010%5E%7B-4%7D%20A)
2) when
, ![V_2=0.31 V](https://tex.z-dn.net/?f=V_2%3D0.31%20V)
Using Ohm's Law, the current is:
![I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A](https://tex.z-dn.net/?f=I_2%3D%5Cfrac%7BV_2%7D%7BR_2%7D%3D%5Cfrac%7B0.31%7D%7B1000%7D%3D3.1%5Ccdot%2010%5E%7B-4%7D%20A)
Now we can rewrite eq.(1) in two forms:
![V_1 = E-I_1 r](https://tex.z-dn.net/?f=V_1%20%3D%20E-I_1%20r)
![V_2=E-I_2 r](https://tex.z-dn.net/?f=V_2%3DE-I_2%20r)
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:
![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](https://tex.z-dn.net/?f=V_1-V_2%3Dr%28I_2-I_1%29%5C%5Cr%3D%5Cfrac%7BV_1-V_2%7D%7BI_2-I_1%7D%3D%5Cfrac%7B0.25-0.31%7D%7B3.1%5Ccdot%2010%5E%7B-4%7D-4.5%5Ccdot%2010%5E%7B-4%7D%7D%3D400%20%5COmega)
b)
To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)
![V=E-Ir](https://tex.z-dn.net/?f=V%3DE-Ir)
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,
is the voltage
is the current
is the internal resistance
Solving for E,
![E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V](https://tex.z-dn.net/?f=E%3DV%2BIr%3D0.25%2B%284.5%5Ccdot%2010%5E%7B-4%7D%29%28400%29%3D0.43%20V)
c)
In this part, we are told that the area of the cell is
![A=4.0 cm^2](https://tex.z-dn.net/?f=A%3D4.0%20cm%5E2)
While the intensity of incoming radiation (the energy received per unit area) is
![Int.=5.5 mW/cm^2](https://tex.z-dn.net/?f=Int.%3D5.5%20mW%2Fcm%5E2)
This means that the power of the incoming radiation is:
![P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W](https://tex.z-dn.net/?f=P%3DInt.%5Ccdot%20A%3D%285.5%29%284.0%29%3D22%20mW%20%3D%200.022%20W)
This is the power in input to the resistor.
The power in output to the resistor can be found by using
![P'=I^2R](https://tex.z-dn.net/?f=P%27%3DI%5E2R)
where:
is the resistance of the resistor
is the current on the resistor (found in part A)
Susbtituting,
![P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W](https://tex.z-dn.net/?f=P%27%3D%283.1%5Ccdot%2010%5E%7B-4%7D%29%5E2%281000%29%3D9.61%5Ccdot%2010%5E%7B-5%7D%20W)
Therefore, the efficiency of the cell in converting light energy to thermal energy is:
![\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20%5Cfrac%7BP%27%7D%7BP%7D%5Ccdot%20100%20%3D%20%5Cfrac%7B9.6%5Ccdot%2010%5E%7B-5%7D%7D%7B0.022%7D%3D0.0044%5Ccdot%20100%20%3D%200.44%5C%25)