take eo1610 19jv k138110 23jk question 4 a solar cell photovoltaic cell has an open circuit volt

(c) The power output of the cell per unit area can be calculated by multiplying the load voltage

Take eo1610 19jv k138110 23jk question 4 a solar cell...

Question 4: A solar cell (photovoltaic cell) has an open circuit voltage value of 0.5 V with a reverse saturation current density of J0 = 1.9×10^-9 A/m^2. The temperature of the cell is 25℃, the load voltage is 0.55 V. The solar irradiation is 900 W/m^2, determine (c) The power output of the cell per unit area considering this load voltage 0.55 V, and the current output density of (b), and (d) The efficiency of the solar cell.

Transcript

00:01 This question we are given here solar cell and the solar cell is photo voltage side the open a circuit that's under a value of 0 .5 volt so here the voltage value v which is equal to 0 .5 volt and the situation current density that j notr you can buy a situation current density j not which is equal to we can write 1 .9 multiple by 10 x to minus 9 m bm meter square and the temperature of the cell we are given here t is equal to 25 celsius and the that is lower voltage that vl which is equal we can write 0 .55 volt 0 .55 volt and the solar radiation is 900 watt per meter square so it's radiation is 900 watt per meter square let's determine the current output density yes so here let's determine the current output density j s that we want to find so here regular given here all the value let's find out that value yes so how to find that first understand that v o c which is equal we can write v multiple by l n of j s divided by g g nod plus one and here v o c which is equal to we can write the formula is joc which is equal to k t divided by e multiple by l n of we can write j s divided by jr plus one now we are substitute value into this formula, so we can add joc which is equal to value is 1 .38 multiple by 10 minus 23, multiple of temperature is 298, divided by electron, there is charge of electron which is 1 .6 multiple of 1019 and l and of 0, we can write j .s divided by value of j0.

01:55 J .0 value here is we can write value of j .nor.

01:59 So here we can write 1 .9 will devere 10 to minus 9 10 to minus 9 1 and now here we just simplify and we can write here answer is the voc which is equal to here we know the voc value is 0 .5 now we just simplify and we can write here value of g s which is equal to answer is 0 .52 6163 mp per meter square so this is the value of g s here now let's write let's find out in second support of this question so in this question we want to find the current density at the load so here we can we can write current density load j s l which is equal we can write j s multiple we can write 1 minus k t divided by q mod by vl which is a substitute of value here so you can buy 0 .563 multiple we can write 1 minus 1 .38 multiple by 10 minus 23 multiplied by 298 divided by 298 divided by here we can write 298 multiple by we can write 1 .6 multiply by 10 minus 90 we can write 0 .55 and now we just simplify and we can write here answer is here we can write the answer is j sl so answer to we can write 0 .501 5566 watt per meter square so this is the value of j sl now let's find out our thoughts are part of this question see since the c we want to find power output so we can write power output which is equal to we know that the values here jvl and vsl which is substituted value here is 0 .55 we can write 0 .501 556 6 6 and we just simply find we can write your answer is so answer is 0 .2727 585 6 6 watt per meter square.

04:12 We are just substitute here which is the answer here, answer of this question, correct answer of this question is...

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