In the circuit shown in fig `E_1 = 3 volts, E_2 = 2volts, E_3 = 1volt and R = r_1 = r_2= r_3= 1 ohm.`

(i) Find the potential difference between the points A and B and the currents through each branch.
(ii) If `r_2` is short circuited and the point A is connected to point B, find the current through `E_1, E_2, E_3` and the resistor R.

Explanation

(a)
`E = ((3)/(1)+(2)/(1)+(1)/(1))/((1)/(1)+(1)/(1)+(1)/(1)) = (6)/(3) = 2 V`
`r= (1)/(3) Omega`
no current in `AX` =, because it is an open circuit, `V_A - V_X = 0`
`V_A - V_B = V_X - V_B = 2V`
Cell of emf `3 V` :
`V_X - V_B = 3 - i_1 xx 1 rArr 2 = 3 - i_1 rArr i_1 = 1 A`
`2 V : V_X - V_B = 2 - i_2 xx 1 rArr 2 = 2 - i_2 = 0`
`1 V : i_1 + i_2 = 1 A`
Hence `V_A - V_B = 2 V`, circuits through branches, `1 A, 0` and `1 A`.

Short-circuit i.e., resistance is replaced by a wire.
From middle branch, `V_X - V_B = 2 V`
Cell of emf ` 3V , V_X - V_B = 3 - i_1 xx 1 rArr 2 = 3 - i_1`
`i_1 = 1 A`, Cell of emf `1 V , V_X - V_B = 1 + i_3 xx 1 rArr 2 = 1 - i_3`
`i_3 = 1 A, X` to `X`, path `XABX`
`V_X - i xx 1 + 2 = V_X rArr i = 2 A`
The distribution of current is an shown.
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