Resistances `R_(1)` and `R_(2)` each 60 are connected in series as shown in figure. The potential difference between `A` and `B` is kept 120 volt. Then what will be tha reading of voltmeter connected between thapoint `C` and `D` if resistance of voltmeter is `120 Omega` ?

Options

A. `48 V`
B. `24 V`
C. `40 V`
D. None

Answer

A

Explanation

(a) `R_(1), R_(2)` and `R_(V)` are in parallel, their equivalent is
`R'_(2) = (R_(2) R_(v))/(R_(2) + R_(V)) = (60 xx 120)/(620 + 120) = 40 Omega`
Now `I = (120)/(R_(1) + R'_(2)) = (120)/(60 + 40) = 1.2 A` reading of voltmeter
`= IR'_(2) = 1.2 xx 40 = 40 V`