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Question
Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.
Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential differences across the equivalent cell.
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Solution
`R_{AB}` = ?
`1/(R_{AB}) = 1/(2r) + 1/(3r) + 1/(6r)`
`1/R_{AB} = 1/r[(3 + 2 + 1)/6]`
`1/R_{AB} = 1/r`
`R_{AB} = r` ⇒ equivalent internal resistant
Equivalent emf of cells in parallel,
`E_{eq} = (E_1/r_1 + E_2/r_2 + E_3/r_3)/(1/r_1 + 1/r_2 + 1/r_3)`
= `(E/(2r) + E/(3r) + E/(6r))/(1/(2r) + 1/(3r) + 1/(6r))`
`E_{eq} = E[(1/(2r) + 1/(3r) + 1/(6r))/(1/(2r) + 1/(3r) + 1/(6r))]`
Eeq = E
Current flowing in the circuit,
I = `E_{eq}/(R + R_{AB})`
I = `E/(R + r)`
Potential drop across E,
V = IReq
V = `(rE)/(R + r)`
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