Resistance `R_1`and `R_2`, each `60 Omega`, are connected in series. The points `A` and `B` is `120 V`. Find the reading of voltmeter connected resistance `r=120Omega`

Answer

B

Explanation

`R_("net")=60+((60)(120))/(60+120)=100 Omega`
`:.i=120/100=1.2A`
Now, reading of voltmeter
`=120-`potential drop across `R_1`
`=120-(60)(1.2)=48V`