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Question
Two identical cells of emf 1.5 V each joined in parallel, supply energy to an external circuit consisting of two resistances of 7 Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.
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Solution
The two cells are connected in parallel. So, the equivalent emf is 1.5 V.
Now, the two resistors are connected in parallel. So, the equivalent resistance is
`1/R_(eq)=1/R+1/R=2/R`
`:.R_(eq)=R/2=7/2=3.5Omega`
The terminal voltage of the cells measured by the voltmeter is 1.4 V.
The net internal resistance of the combination of cells is
`r_(eq)=((varepsilon-V)/V)R`
`:.r_(eq)=(1.5-1.4)/1.4xx3.5=0.1/1.4xx3.5=0.25 Omega`
Now, the individual internal resistors are connected in parallel. So, the individual internal resistances is
`r_(eq)=(r')/2`
∴ r' = 2req = 2 x 0.25 = 0.5Ω
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