A solar cell generates a potential difference of 5.62 V when a 1.20 k Ohm extern

Question

A solar cell generates a potential difference of 5.62 V when a 1.20 k Ohm external resistance is connected across it, i.e., from one terminal of the cell to the other terminal of it, and a potential difference of 6.84 V when a 2.90 k Ohm external resistance is used. What is the internal resistance r_i of the solar cell? (Give your answer in the form of "abc" Ohm.) 8.25 Times 10^(2) Ohm. 7.67 Times 10^(2) Ohm. 4.67 Times 10^(3) Ohm. 5.93 Times 10^(2) Ohm. 3.93 Times 10^(3) Ohm. What is the emf sigma (V) of the solar cell? 9.93V. 7.67J. 7.07V. 5.93J. 8.08V. For the circuit diagram as Figure 4 of Assignment 7, sigma = 15 kV; R = 4.5 k Ohm, and C = 7800.0 mu F.

Explanation / Answer

Let the emf of the solar cell be E and the output voltage be V.

V=E-Ir

For R=1.2 Kohms

I=V/R = 5.62/1.2*103 = 4.6833*10-3 A

5.62=E-(4.6833*10-3)r

E-(4.6833*10-3)r=5.62---------------------1

For R=2.9 Kohms

I=6.84/2900 =2.3586*10-3 A

E-(2.3586*10-3)r =6.84--------------------2

Solving 1 and 2 we get

E=8.08 Volt

r=525 ohms

a)

Answer is a)5.25*102 ohms

b)

Answer is e)E=8.08 V

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