A circuit consists of three bateries of emf `E_(1) = 1V, E_(2)=2V` and `E_(3)=3V` and internal resistance `1Omega, 2Omega` and `1Omega` respectively which are connected in parallel as shown in 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

`1Omega, 2Omega` and `1Omega` are in parallel.

So, the required internal resistance.
`(1)/(r )=(1)/(r_(1))+(1)/(r_(2))+(1)/(r_(3))rArr (1)/(r )=(1)/(1)+(1)/(2)+(1)/(1)`
`(1)/(r )=(2+1+2)/(2)rArr r=(2)/(5)Omega`
The potential difference between points P and Q
`E_("diff")=((E_(1))/(r_(1))+(E_(2))/(r_(2))+(E_(3))/(r_(3)))/(1//r)=((1)/(1)+(2)/(2)+(3)/(1))/(5//2)`
`= ((2+2+6)/(2))/(5//2)=(10//2)/(5//2)=(5)/(5)xx2=2V`