Continuing the previous problem determine the power dissipat

Continuing the previous problem, determine the power dissipated in R_2. (Give your answer in the form of \"a. bc Times 10 (x) W) A solar cell generates a potential difference of 1.62 V when a 1.50 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 1.84 V when a 2.70 k Ohm external resistance is used. What is the internal resistance r_j of the solar cell? (Give your answer in the form of \"abc\" Ohm.)

Solution

Let the EMF of the solar cell be \"E\" and Output Voltage be \"V\"

V=E-Ir-------------1

where r =internal resistance

I=Current flowing through the resistor

Case 1:V=1.62 V and R=1.5 Kohms

I=V/R =1.62/1500 =1.08*10^-3 A

=>1.62 =E-(1.08*10^-3)r

E-0.00108r=1.62-----------------------------------2

Case 2:V=1.84 Volts and R=2.7 Kohms

I=1.84/2700 =6.81*10^-4

=>1.84=E-(6.81*10^-4)r

E-0.000681r=1.84--------------------3

solving 2 and 3 we get

E=2.22 Volts

r=552 ohms

So internal resistance of the solar cell is

r_i=552 ohms