Using Kirchhoff's rules, calculate the potential difference between B and D in the circuit diagram as shown in Fig.

Explanation

Let currents flowing through different branches of the circuit be as shown in Fig. Then, as per Kirchhoff.s first rule `I = I_1 + I_2`
Applying Kirchhoff.s second rule to mesh BADB, we have
`I_1. 2 - I_1.1 + I_2.2 = 2 - 1 `
` rArr I_1 + 2I_2 = 1`
Similarly for mesh DCBD, we have
`I.3 - I.1 - I_2.2 = 3 - 1 rArr 4I - 2I_2 = 2`
`4(I_1 + I_2) - 2I_2 = 2 " or 4I_1 + 2I_2 = 2` ...(ii)
Solving (i) and (ii), we get `I_1 = I_2 = 1/3A`
` therefore ` Potential difference between B and D, `V_(BD) = I_2 xx2= 5 xx2=+ 2/3 `