Answer to Question #280043 in Electric Circuits for aly

Question

1. Eight groups of cells are in parallel. Each group consists of  four cells, each of emf 4 V and internal resistance 0.5 Ώ, In series. Determine the power supplied to 4 Ώ resistor by this battery system

Answer 1

Total emf of one group will be, $e = 4+4+4+4 = 16 V$

Total resistance of one group, $r = 0.5+0.5+0.5+0.5 = 2 \Omega$

Total emf of eight groups, $e = 16 V$ as voltage remains same in parallel.

Total resistance of eight groups, $r' = \frac{2}{8} = 0.25 \Omega$

Total resistance of the circuit will be, $R = r' + 4 = 4+0.25 = 4.25\Omega$

Total current in the circuit is, $I = \frac{e}{R} = \frac{16}{4.25} = 3.77 A$

Power dissipated in 4 ohm resistance, $P = 4I^2 = 4 \times (3.77)^2 = 56.69 W$