Question

For the circuit shown, with R₁ = 1.0Ω, R₂ = 2.0Ω, E₁ = 2 V and E₂ = E₃ = 4 V, the potential difference between the points ‘a’ and ‘b’ is approximately (in V):
(A) 3.3 (B) 2.3
(C) 3.7 (D) 2.7

Answer 1

Answer:

sorry don't know the answer

Answer 2

The potentianl difference between the points a and b in the circuit is 3.3V,

• Let i1 and i2 be the currents in the two loops.

Consider the middle branch,

• Using Kirchoff's Voltage Law,

Let Va - Vb = V, to be found.

• Vb + 4 - (i2 - i1)R2 - Va =0 ==>
• V = 4 + 2(i1 - i2) --(1)

Using Krichoff's Voltage law in loop1

• Vb - R1xi1  + E1 - R1xi1 - Va = 0 ==>
• V = 2 - 2xi1 ==> i1 = (2- V)/2 ---(2)

Using Kirchoff's Voltage Law in loop2

• Va - i2xR1 - 4 - i2xR1 -Vb =0 ==>
• V = 4 + 2xi2 ==> i2 = (V - 4)/2 --(3)

Substituting (2) and (3) in (1) ==>

• V = 4 + 2( 1 - V/2 - V/2 + 2) = 4 + 2(3 - V) = 4 + 6 - 2V==>
• 3V = 10 ==>
• V = 3.3V

Hence the potential difference between a and b is 3.3V