Find the potential difference across each of the four batteries `B_1, B_2, B_3` and `B_4` as shown in the figure.

Explanation

Across `B_1` This battery does not make any closed circuit
`:. i=0` or `V=E=4 "volt"`
Across `B_2` This battery is short circuited. Therefore
`V=0`
Across `B_3` and `B_4` A current in anti clockwise direction flows in the closed loop `abcda`. This current is
`i=("net emf")/("net resistance")=(10-5)/(1+2+2)`
=1A
Now, current flows through `B_3` in normal direction. Hence,
`V=E-ir=10-1xx1=9 "volt"`
From `B_2` current flows is opposite direction.Hence,
`V=E+ir=5+1xx2=7 "volt"`