Calculate the voltage across AB terminals in the given circuit ,

Options

A. `3/8 V`
B. `8/3 V`
C. `3/2 V`
D. `2/3 V`

Answer

B

Explanation

Given, circuit diagram is shown in the figure,
`therefore ` Current in branch AB,
`V_(AB) = I_(AB) xx R_(AB) = 8/30 xx 10 = 8/3 V` If I and I be the loop currents in loop (1) and loop (2), respectively then

By KVL in loop (1),
`- 5 + 10I_1 + 10 (I_1 - I_2) = 0`
` 20 I_1 -10I_2 = 5 " " ...(i)`
By KVL in loop (2).
`10I_2 + 3 + 10 (I_2 - I_1) = 0`
`-10I_1 + 20I_2 = - 3`
`- 20I_1 + 40I_2 = - 6 " " ...(ii)`
Solving Eqs. (i) and (ii), we get
` 30I_2 = -1`
`I_2 = (-1)/(30) A`
From Eq. (i), `20I_1 - 10 ( (-1)/(30) ) = 5 rArr I_1 = 7/30 A`
` therefore ` Current in branch AB, `I_(AB) = I_1 - I_2 = 7/30 - (-1)/(30)`
`I_(AB) = 8/30 A`
` therefore ` Voltage across a terminal AB ,
`V_(AB) = I_(AB) xx R_(AB) = 8/30 xx 10 = 8/3 V`