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.
Answers
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.