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# AS physics-electricity watch

1. Does anyone know how to do these:

A cell of e.m.f. 1.5 V and internal resistance 2.5 Ohms is connected to a resistor of 3.5 Ohms. Find the current in the circuit.

When a battery of e.m.f. 3 V is connected to a resistor, the current taken from the battery is 2 A. If the internal resistance of the battery is 1 Ohm, find the value of the resistor.
2. 1) I = V/R = 1.5/(2.5+3.5) = 0.25 A

2) (1+r) = V/i = 3/2 = 1.5 => r = 1.5 - 1 = 0.5 Ohm

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