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Let two cells of e.m.f.'s E_1 and E_2 be connected in parallel in a circuit . Let r_1 and r_2 be the internal resistances of the cells . Find the value of the current

Answer»

Solution :Resultant p.d. due to `E_1 and E_2 = E_1 - E_2`
SUM of INTERNAL RESISTANCE = `r_1 + r_2`
`therefore ` current `I =(E_1 - E_2)/((r_1 +r_2))`


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