Question

12 cells each emf 1.5V and resistance 0.5ohm are arranged in m rows each containing n cells connected in series. calculate values of m & n for which the combination would send max current through external resistance of 1.5ohm.

Answer 1

Given: 12 cells each emf 1.5 V and resistance 0.5 ohm are arranged in m rows each containing n cells connected in series.

To find: Calculate values of m & n for which the combination would send max current through external resistance of 1.5 ohm.

Solution:

• Now we have given that the circuit consists of m rows & n cells.

• Then the effective emf of each branch will be:

                ∑(row) = n∑ = 1.5n V

• Now the internal resistance will be:

                r(row) = nr = 0.5n ohm

• There are m rows so effective emf and internal resistance will be:

                1/r(eq) = m / nr

                r(eq) = nr/m

                r(eq) = 0.5 x (n/m) ohm

                ∑(eq) / r(eq) = mn∑/nr

                ∑(eq) = m∑ x r(eq)

                ∑(eq) = m x 1.5 x 0.5n/m

                ∑(eq) = 0.75n V

• So the current in resistor of 1.5 ohm will be:

                I = ∑(eq) / (R + r(eq))

                I = 0.75n / (1.5 +0.5n/m)

                I = 0.75nm / 1.5m + 0.5n

                I = 9 /  1.5m + 0.5n

• For I to be max, the derivative will be equal to 0.

• So:

                d/dm (1.5m + 0.5n) = 0

• Putting value of n, we get:

                d/dm (1.5m + 0.5(12/m)) = 0

                d/dm ((1.5m^2 +6) /m) = 0

                (1.5m^2 +6) - m(3m) / m^2 = 0

                (1.5m^2 +6) - 3m^2 = 0

                - 1.5m^2 + 6 = 0

                1.5m^2 = 6

                m^2 = 4

                m = 2

                So therefore n = 6

Answer:

              So the value of m is 2 and n is 6.