1.

In the circuit shown in the given figure, the resistances R_(1) and R_(2) are respectively

Answer»

`14Omega and 40Omega`
`40Omega and 44Omega`
`40Omega and 30Omega`
`14Omega and 30Omega`

Solution :Potential difference ACROSS `20Omega=20xx1=20V ="potential difference across" R_(2).`
Current in `R_(2)=0.5A`
`THEREFORE R_(2)=(20V)/(0.5A)=40Omega=40Omega`
Potential difference across `R_(1)=69V-20V=49V`
Current in `R_(1)=0.5A+(20)/(10)+1A=3.5A`
`therefore R_(1)=(49)/(3.5)=14Omega`


Discussion

No Comment Found

Related InterviewSolutions