Shown two ideal voltmeters and an ammeter, which are connected across the various circuit elements. If the voltmeter connected across `9Omega` resistance reads 4.5 V, then answer the following problems.

The current through `12Omega` resistance is

Options

A. `0.1A`
B. `0.75A`
C. `0.5A`
D. `1.25A`

Answer

C

Explanation

`I_(1) = (4.5)/(9) = (1)/(2)A`
Potential difference across AB:
`4.5 +3I_(1) = 6I_(2) or I_(2) = 1 A, I = I_(1) + I_(2) = 1.5A`
Equivalent resistance between C and D: `R_(1) =4Omega`
Potential difference across `CD : R_(1) I = 4xx1.5 =6V`
Current through `12Omega, I_(3) = (6)/(12) = (1)/(2)A`
Potential difference across AD is
`6I_(2) +2I +12I_(3) = 6xx1+2xx1.5+12 xx0.5 =15V`
This will be equal to the reading of voltmeter across `20Omega`
`I_(4) = (15)/(20) = 0.75A`
Reading of the ammeter is `I +I_(4) =1.5 + 0.75 = 2.25A`