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In the shown circuit...

In the shown circuit

A

The power developed in `4 Omega` is maximum

B

The power developed in `2 Omega` is more as compared to `3 Omega`

C

The power developed in `12 Omega` depends on emf E if battery

D

The current flowing through `6 Omega` resistance is `(5E)/(48)A`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D

`V_(1) = (5)/(8) E V_(2) = (3)/(8) E`
`P_(4 Omega) = ((5)/(8) V)^(2) (1)/(4)` maximum
and `P_(2 Omega)=((3)/(8)V)^(2)(1)/(2) rArrP_(2 Omega)=((3V)/(8))^(2)(1)/(3)`
Here `P_(4 Omega) gt P_(2 Omega)`
current in `6 Omega =(5E)/(8)(1)/(6)rArr i=(5E)/(48)A`
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