1.

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

Answer»

`(1)/(4)W`
`(1)/(2)W`
2W
4W

Solution :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)`


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