A solar cell generates a potential difference of 0.10Vwhen a500 resistor is connected across it, and a potential difference of 0.15Vwhen a 1000resistor is substituted.

(a) What is the internal resistance?

(b) What is the emf of the solar cell?

(c) The area of the cell is5.0cm2 , and the rate per unit area at which it receives energy from light is2.0mW/cm2 .What is the efficiency of the cell for converting light energy to thermal energy in the1000 external resistor?

Short Answer

Expert verified

a) The internal resistance of the solar cell is .1.0×103Ω

b) The emf of the solar cell is0.30V.

c) The efficiency of the solar cell is0.23%

Step by step solution

01

Step 1: Given

V=0.10VR=500ΩV'=0.15VR'=1000ΩA=5.0cm2=5×10-4m2PinputA=2.0mWcm2=20Wm2

02

Determining the concept

Write two expressions for emf of the solar cell for both the voltages and the resistances using Ohm’s law. Equating them can give the internal resistance. Inserting the value of internal resistance in one of the equation will give emf of the cell. Then from the rate of the energy received per unit area, find the efficiency using the corresponding formula.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points

Formulae are as follow:

I=VR

Where, I is current, V is voltage, R is resistance.

03

(a) determining the internal resistance of the solar cell

Let the internal resistance of the battery be r and emf beε.

When the resistance R is in the circuit, emf is

ε=V+Ir

But, according to the Ohm’s law,

I=VRε=0.10+0.10500rε=0.10+2×104r..1)

When resistance R’ is in the circuit, emf is,

ε=V'+Ir

But, according to the Ohm’s law,

I=V'R'ε=0.15+0.151000rε=0.15+1.5×104r.1)

Equating 1) and 2) gives,

r=1.0×103Ω

Hence,the internal resistance of the solar cell is 1.0×103Ω.

04

(b) determining the emf of the solar cell

Inserting value of r in equation 1 yields,

ε=0.10+2×104(1.0×103)

ε=0.30V

Hence, the emf of the solar cell is0.30V.

05

(c) Determining the efficiency of the solar cell

Efficiency of the solar cell is,

η=PoutputPinput=V2RPinputη=(0.15)2100020×5×104η=2.25×103~0.23%

Hence, the efficiency of the solar cell is0.23%

Therefore, the internal resistance, emf and the efficiency of the solar cell from the potential difference, the resistance across it and the power received can be determined using Ohm’s law and the corresponding formulae.

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