In the network shown the potential difference between `A` and `B` is `(R = r_(1) = r_(2) = r_(3) = 1 Omega, E_(1) = 3V, E_(2) = 2V, E_(3) = 1V`)

Options

A. `1 V`
B. `2 V`
C. `3 V`
D. `4 V`

Answer

B

Explanation

(b) No current through `R`, so potential difference across
`AB` is `V = ((E_(1))/(r_(1)) + (E_(2))/(r_(2)) + (E_(3))/(r_(3)))/((1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3))) = ((3)/(1) + (2)/(1) + (1)/(1))/((1)/(1) + (1)/(1) + (1)/(1)) = 2 V`