How Do You Calculate the Internal Resistance and EMF of a Solar Cell?

[incubusguy88]

Someone please help me out with this problem, I can solve for the current but from there i get stuck...

A solar cell generates a potential difference of 0.10 V when a 600 resistor is connected across it, and a potential difference of 0.15 V when a 1150 resistor is substituted. What are
(a) the internal resistance and


(b) the emf of the solar cell?


(c) The area of the cell is 2.0 cm2, and the rate per unit area at which it receives energy from light is 2.0 mW/cm2. What is the efficiency of the cell for converting the light energy to thermal energy in the 1150 external resistor?

 

[turin]

Originally posted by incubusguy88
What are
(a) the internal resistance and

(b) the emf of the solar cell?

The best thing to do is solve these two simultaneously (2 eqs. and 2 unkwns.). Can you please describe how you are modelling your circuit?

Originally posted by incubusguy88
What is the efficiency of the cell for converting the light energy to thermal energy in the 1150 external resistor?

You need to know the definition of efficiency, and how to find the power delivered to the load resistor. Do you know these two things?

 

[M98Ranger]

Sure, I can help you with this DC circuits problem. First, let's start by writing out the given information:

Potential difference (V) = 0.10 V and 0.15 V
Resistance (R) = 600 Ω and 1150 Ω
Area (A) = 2.0 cm2
Rate of energy (P) = 2.0 mW/cm2

(a) To solve for the internal resistance (r) of the solar cell, we can use the formula V = EMF - Ir, where V is the potential difference, EMF is the electromotive force (emf) of the solar cell, and I is the current. We know the potential differences for both resistors, so we can set up two equations:

0.10 V = EMF - (I)(600 Ω)
0.15 V = EMF - (I)(1150 Ω)

We can then solve for I in both equations and set them equal to each other to eliminate I:

0.10 V = EMF - (0.10 V/600 Ω)(600 Ω)
0.15 V = EMF - (0.15 V/1150 Ω)(1150 Ω)
0.10 V/600 Ω = 0.15 V/1150 Ω
0.0001667 A = 0.0001304 A
EMF = 0.104 V

Now, we can plug in this value for EMF into either equation to solve for I:

0.10 V = 0.104 V - (I)(600 Ω)
0.0004 A = I

Finally, we can plug in this value for I into the formula for internal resistance:

r = (0.10 V)/(0.0004 A)
r = 250 Ω

So, the internal resistance of the solar cell is 250 Ω.

(b) To solve for the emf of the solar cell, we can simply plug in the values we found for EMF and internal resistance into the formula:

EMF = V + Ir
EMF = 0.10 V + (0.0004 A)(250 Ω)
EMF = 0.20 V

So, the emf of the solar cell is 0.20 V.

(c) To find the efficiency of the solar cell, we can use the