Two `2Omega` resistances are connected in parallel in the circuit X and in series in circuit Y. The batteries in the two circuits are identical and have zero internal resistance. Assume that the energy transferred to resistor A in circuit X within a certain time is W. The energy transferred to resistor B in circuit Y in the same time will be

Options

A. `(1)/(4)W`
B. `(1)/(2)W`
C. 2W
D. 4W

Answer

A

Explanation

In a circuit X, both the resistance are in parallel, Therefore V is the same and, Power (energy transferred in unit time) is `(V^(2))/(2)=W`
In a circuit Y, both resistance are in series.
Therefore, `V_(B)+V_(B)'=V or V_(B)=V//2`
In a circuit Y, power supplied to `B=((V//2)^(2))/(2)=(V^(2))/(8)=(W)/(4)`