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(A) The total number of electroplaques is $N = 5000$. The number of rows is $P = 100$. The number of electroplaques in each row is $S = N/P = 5000/100

Vedclass · Accepted Answer

$1.5$

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(A) The total number of electroplaques is $N = 5000$. The number of rows is $P = 100$. The number of electroplaques in each row is $S = N/P = 5000/100 = 50$.Each electroplaque has $emf$ $e = 0.15\ V$ and internal resistance $r = 0.25\ \Omega$.The total $emf$ of one row is $E_{row} = S 	imes e = 50 	imes 0.15 = 7.5\ V$.Since all rows are in parallel,the total $emf$ of the arrangement is $E_{total} = 7.5\ V$.The internal resistance of one row is $r_{row} = S 	imes r = 50 	imes 0.25 = 12.5\ \Omega$.The total internal resistance of $P$ rows in parallel is $r_{total} = r_{row} / P = 12.5 / 100 = 0.125\ \Omega$.The external resistance of the water is $R = 500\ \Omega$.The total current $I$ in the circuit is given by $I = E_{total} / (R + r_{total})$.$I = 7.5 / (500 + 0.125) = 7.5 / 500.125 \approx 0.01499\ A \approx 0.015\ A$.Wait,checking the calculation: $S = 5000/100 = 50$. $E_{total} = 50 	imes 0.15 = 7.5\ V$. $r_{total} = (50 	imes 0.25) / 100 = 0.125\ \Omega$. $I = 7.5 / 500.125 \approx 0.015\ A$. Looking at the options,there might be a misinterpretation of the provided image text. If $5000$ is per row,then $S=5000$. $E_{total} = 5000 	imes 0.15 = 750\ V$. $r_{total} = (5000 	imes 0.25) / 100 = 12.5\ \Omega$. Then $I = 750 / (500 + 12.5) = 750 / 512.5 \approx 1.46\ A \approx 1.5\ A$. This matches option $A$.

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