A new battery has an emf of 2,99 V. When a wire of negligible resistance is connected between the terminals of the battery, a current of 9,09 A is produced. Find the internal resistance of the battery.
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The battery is 3.0v, internal resistance, Ri = 0.5 Ohms. A 4.0 Ohm resistor is connected as a load.
The total resistance around the loop is Rt = 0.5 + 4.0 = 4.5 Ohms.
The loop current is I = 3.0v/4.5 Ohms = 2/3 Amp
The terminal voltage is Vt = 3.0v - 0.5 Ohms x 2/3 Amp = 3.0v - 0.333v = 2.667v.
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