Fing the reading of the ideal ammeter connected in the given circuit. Assume that the cells have negligible internal resistance.

Options

A. `0.8 A`
B. `0.25 A`
C. `1.95 A`
D. `0.05 A`

Answer

C

Explanation

(c ) In loop `ghjig`, `-2I_(1) - 6 = 0 implies I_(1) = - 3A`
In loop `efhge`, `- 4 I_(2) + 2 I_(1) + 8 = 0 implies I_(2) = (1)/(2) A`
In loop `cdfec`, `- 8 I_(3) + 4 I_(2) - 4 = 0 implies I_(3) = - (1)/(4) A`
In loop `abdca`, `- 10 I_(4) + 8 I_(3) + 10 = 0 implies I_(4) = (4)/(5) A`
Current through ammeter `= I_(1) + I_(2) + I_(3) + I_(4)`
`= (-60 + 10 - 5 + 16)/(20) = (39)/(20) A = 1.95 A`