Four identical cells- each of emf 2V are joined in parallel providing supply of current to external circuit consisting of two 15- resistors joined in parallel- The terminal voltage of the cells as read by an ideal voltmeter is 1-6 volt- Calculate the internal resistance of each cell-

#10; External resistance is 15 Ω 2 (resistors) = 7.5 Ω . #10;Current = 1.6 volts 7.5 Ω = 0.2133 Amps . #10; 0.2133 Amps × R ...

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Standard XII

Physics

Non-Ideal Battery

Four identica...

Question

Four identical cells, each of emf $2 V$ are joined in parallel providing supply of current to external circuit consisting of two $15 Ω$ resistors joined in parallel. The terminal voltage of the cells as read by an ideal voltmeter is $1.6$ volt. Calculate the internal resistance of each cell.

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Solution

External resistance is $\frac{15 Ω}{2 (r e s i s t o r s)} = 7.5 Ω$ .
Current $= \frac{1.6 v o l t s}{7.5 Ω} = 0.2133 A m p s$ .
$0.2133 A m p s \times R = 2.0 - 1.6 = 0.4 v o l t s$ .
$R = 0.4 v o l t s 0.2133 A m p s = 1.875 Ω$ .

$1.875 Ω$ is the internal resistance of the $4$ batteries in parallel.
The internal resistance of a single cell is $1.875 Ω$ time $4$ (cells) = $7.5 Ω$ .

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Four identical cells, each of emf 2V are joined in parallel providing supply of current to external circuit consisting of two 15Ω resistors joined in parallel. The terminal voltage of the cells as read by an ideal voltmeter is 1.6 volt. Calculate the internal resistance of each cell.

External resistance is 15Ω2(resistors)=7.5Ω. Current =1.6volts7.5Ω=0.2133Amps. 0.2133Amps×R=2.0−1.6=0.4volts. R=0.4volts0.2133Amps=1.875Ω. 1.875Ω is the internal resistance of the 4 batteries in parallel. The internal resistance of a single cell is 1.875Ω time 4 (cells) = 7.5Ω.

Four identical cells, each of emf $2 V$ are joined in parallel providing supply of current to external circuit consisting of two $15 Ω$ resistors joined in parallel. The terminal voltage of the cells as read by an ideal voltmeter is $1.6$ volt. Calculate the internal resistance of each cell.