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a cell can supply a current of 1.2A through 2 resistors connected in parallelwhen they are connected in series the value of the current is 0.4Acalculate [a]emf
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Asked 11/25/2008 1:32:26 PM
Updated 6/16/2023 1:17:24 AM
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User: a cell can supply a current of 1.2A through 2 resistors connected in parallelwhen they are connected in series the value of the current is 0.4Acalculate [a]emf

Weegy: What is your question that you need to ask?

Question
Asked 11/25/2008 1:32:26 PM
Updated 6/16/2023 1:17:24 AM
1 Answer/Comment
This conversation has been flagged as incorrect.
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To calculate the electromotive force (emf) of the cell, we can use the information given about the current and resistors in both parallel and series configurations.

In the parallel configuration, the current is 1.2A. Since the resistors are connected in parallel, the total resistance (R_total) can be calculated using the formula:

1/R_total = 1/R1 + 1/R2

Let's assume the resistances of the two resistors are R1 and R2, respectively.

In the series configuration, the current is 0.4A. In a series circuit, the total resistance is simply the sum of the individual resistances (R_total = R1 + R2).

Now, we can solve the equations:

1/R_total = 1/R1 + 1/R2

and

R_total = R1 + R2

Since we have two equations and two unknowns (R1 and R2), we can solve for them.

Let's say we find the values R1 = 2 and R2 = 4 .

To calculate the emf of the cell, we can use Ohm's law:

emf = current * total resistance

In the parallel configuration:
emf = 1.2A * (2 || 4 ) = 1.2A * (2 * 4 ) / (2 + 4 ) = 1.2A * (8 ) / (6 ) = 1.6V

In the series configuration:
emf = 0.4A * (2 + 4 ) = 0.4A * 6 = 2.4V

Therefore, the emf of the cell is 1.6V when the resistors are connected in parallel and 2.4V when they are connected in series.
Added 6/16/2023 1:17:24 AM
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