(a) Find the current in `5 Omega` and voltage across `5 Omega`.

(b) Find the current in the `15 Omega` resistor.

( c) Calculate the potential difference between `A` and `B`.

(d) Find the current flowing in `10 Omega` and the potential difference between `A` and `B` .
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Explanation

(a)
`i= (30)/(5 + 5) = 3 A`
`V_A - V_B = i xx 5 = 3 xx 5 = 15 V`
(b)
`E = ((12)/(6)+(3)/(3))/((1)/(6) +(1)/(3)) = 24 V`
`r = (6 xx 3)/(6 + 3) = 2 Omega`

`i = (24)/(2 + 6) = 3 A`
`V_A - V_B = i xx 6 = 3 xx 6 = 18 V`

`V_A - V_B = i_0(15 + 3)`
`18 = i_0 xx 18`
`i_0 = 1 A`
( c)
`E = ((6)/(6)-(6)/(6)+(4.5)/(3))/((1)/(6)+ (1)/(6) +(1)/(3)) = (4.5)/(2) = 2.25 V`
`(1)/(r)=(1)/(6)+(1)/(6)+(1)/(3) = (1+1+2)/(6) = (1)/(1.5) rArr r = 1.5 Omega`

`i = (2.25)/(1.5 + 3) = (1)/(2) A`
`V_A - V_B = i xx 3 = (1)/(2) xx 3 = 1.5 V`
(d)
`E = ((1.5)/(10)+(2)/(20)-(2.5)/(30))/((1)/(10)+(1)/(20)+(1)/(30)) = ((9 + 6 -5)/(60))/((6 + 3 +2)/(60)) = (10)/(11) V`
Equivalent cell is open circuit hence no need to find equivalent resistance `r`.
`V_A - V_B = (10)/(11) V`
Cell of emf `1.5 V`,
`V_A - V_B = 1.5 - i_0 xx 10`
`(10)/(11) = 1.5 - 10 i_0`
`10 i_0 = 1.5 -(10)/(11)`
`i_0 ~= 0.06 A`.