Three resistances `2Omega, 4Omega, 5Omega` are combined in series and this combination is connected to a battery of 12 V emf and negligible internal resistance. The potential drop across these resistances are

Options

A. `(5,45,4.36,2.18)V`
B. `(2,18,5.45,4.36)V`
C. `(4.36,2.18,5.45)V`
D. `(2.18,4.36,5.45)V`

Answer

D

Explanation

Let current in the circuit is I. Then total resistance in the circuit.
`R=R_(1)+R_(2)+R_(3)=2+4+5=11Omega`
`therefore V=IR therefore I=(V)/(R)=(12)/(11)A`
The potential drop across `2 Omega` resistance
`V_(1)=IR_(1)=(12)/(11)xx2=2.18V`
The potential drop across `4Omega` resistance
`V_(2)=IR_(2)=(12)/(11)xx4=4.36V`
The potential drop across `5Omega` resistance
`V_(3)=IR_(3)=(12)/(11)xx5=5.45V`
`"Hence" (V_(1),V_(2),V_(3))=(2.18,4.36,5.45)V`