Question
Four identical cells, each having an emf of 1.8 V,
are linked in parallel and connected to an external network made of two resistors of 10 Ω each arranged in parallel. If the potential difference across the output terminals is measured to be 1.5 V by an ideal voltmeter, what is the internal resistance of each cell?Solution
When multiple identical cells are joined in parallel, the net emf remains the same as that of a single cell. Hence,
Total emf (E) = 1.8 V Let r be the internal resistance of each cell. Since the cells are in parallel, their effective internal resistance is reduced: 
Now, calculate the combined resistance of the external circuit: Two 10 Ω resistors in parallel: 
Using Ohm’s law: 
The potential drop inside the battery system due to internal resistance is: 
45.1298% of (14.032 - 75.98) + 27.87% of √40001 = 449.98% of 24.098 + ?
5.25 x 4.09 + 3.99 x 9.67 + 6.01 x 14.88 = ?
24.99% of 1619.78 + (1259.84 ÷ 12.24) = ? × 16.98
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
54.97% of 620.08 + 16.13 × 11.11 – 829.91 ÷ 5 = ?
37.06% of 783.45 + 2125% of 51.89 = ?
Find the approximate value of Question mark(?). No need to find the exact value.
24.95 × (36.06 ÷ 6) + 74.95% of 159.89 – √(143.94) × 2....
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
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