a cell of e.m.f. 1.5V and internal resistance 1.0\(\Omega\) is connected to two resistor of resistance 2.0\(\Omega\) and 3.0\(\Omega\) in series. Calculate ...
Question 1 Report
a cell of e.m.f. 1.5V and internal resistance 1.0\(\Omega\) is connected to two resistor of resistance 2.0\(\Omega\) and 3.0\(\Omega\) in series. Calculate the current through the resistors
Answer Details
The total resistance in the circuit is the sum of the individual resistances:
R = 2.0\(\Omega\) + 3.0\(\Omega\) = 5.0\(\Omega\)
The current in the circuit is given by Ohm's Law:
I = V / R, where V is the voltage across the circuit, which is equal to the emf of the cell, E.
Therefore, I = E / (R + r), where r is the internal resistance of the cell.
Substituting the given values, we get:
I = 1.5V / (5.0\(\Omega\) + 1.0\(\Omega\)) = 0.25A
Therefore, the current through the resistors is 0.25A.
Hence, the answer is (a) 0.25A.
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