A circuit consists of three batteries of emf `E_1` = 1 V, `E_2= 2 V and E_3 = 3 V` and internal resistances `1 Omega ,2 Omega and 1 Omega` respectively which are connected in parallel as shown in the figure. The potential difference between points P and Q is

Options

A. `1.0 V`
B. `2.0 V`
C. `2.2 V`
D. `3.0 V`

Answer

B

Explanation

Here ` E_1 = 1 V, E_2 =2 V ,E_3 =3V`
` r_1 =1 Omega , r_2 = 2 Omega , r_3 = 1 Omega`
the effective emf of the circuit is
` E_("eff")=((E_1)/( r_1)+(E_2)/(r_2) +(E_3)/(r_3))/((1)/(r_1) +(1)/(r_2) +(1)/(r_3) )= ((1)/(1) +(2)/(2) +3/1)/(1/1 +1/2 + 1/1 )= ( (10)/(2) )/(5/2) = 10/5 =2 V `
` therefore ` potential difference between points P and Q ` = E_("eff") = 2V `