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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. - CBSE (Science) Class 12 - Physics

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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.

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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Solution 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. Concept: Cells, Emf, Internal Resistance.
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