Answer to Question #263259 in Electrical Engineering for Jose ygot

Question

Five batteries are connected in parallel to supply a certain load. Each battery consist of 12 cells in series. Each cell has an emf of 2V and the internal resistance of each cell is 0.15Ω. What should be the resistance of the load so that the total power consumed is 384 watts?

Answer 1

As all cells are in series, therefore, total voltage in one branch = 2 * 12 = 24V

Also all batteries are connected in parallel across the load, therefore, net voltage applied across the load = 24 V

Total Power = 384 W

Internal Res. of each cell = 0.15 Ohms

Therefore, total res. in one branch = 0.15 * 12 = 1.8 Ohms.

Now combination of 1.8 Ohms resistance are in parallel of five no. of branches.

Therefore The load resistance RL is in series with (1.8 || 1.8 || 1.8 || 1.8 || 1.8)

Net Eq. Res. = RL + (1.8/5) = RL + 0.36

Given Power P = 384 = $\frac{V^2}{R}$

$R = \frac{24 * 24}{384} = 1.5 Ohms$

R = RL + 0.36 = 1.5

RL = 1.5 - 0.36 = 1.14 Ohms

Therefore, a load RL = 1.14 Ohms will draw a power of 384 Watts in above circuit.