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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

#### 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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