Two resistors of 5 `Omega and 10 Omega` are connected in parallel to a battery of emf 16 V and a current of 4 A is drawn from the battery . How will you explain the given magnitude of current ? Do you see any discrepancy in the given magnitude of current ?

Explanation

As the resistance of `5Omega and 10Omega` are connected in parallel, their resultant resistance is:
`(1)/(R ) =(1)/(5) + (1)/(10)`
`(1)/(R ) = (3)/(10)`
`R= (10)/(3) Omega`
emf of battery, V=16V
`I= (V)/(R ) = (16xx3)/(10) =4.8A`
There must be an internal resistance of the battery due to which the actual magnitude of current in the circuit is reduced from 4.8 to 4A.