Maximum power in a 0.5`Omega` resistance connected with two batteries of 2V emf and 1`Omega` internal resistance in parallel, is ........ .

Options

A. `(8)/(9) ` W
B. 1.28 W
C. 2.0 W
D. 3.2 W

Answer

C

Explanation

2.0 W
Equivalent resistance of internal resistance
`r. = (1 xx 1)/(1 + 1) = (1)/(2) = 0.5 Omega`
Total resistance of circuit = 0.5 + 0.5 = ` 1.0 Omega`

`therefore` Current flowing through the circuit
`I = (E)/(R) = (2)/(1) = 2 `A
`therefore ` As 2A current is flowing through the 0.5 `Omega` resistor, so P `= I^(2) R = 4 x 0.5 = 2.0 ` W