EDS of batteries 4 5 V, wire resistance 0.01 Ohm. Is it possible to get a current of hundreds of amperes by closing the circuit.

Given: εi (battery EMF) = 4.5 V; R (wire resistance) = 0.01 ohm.

1) Approximate calculation: EMF of 4.5 V has a 3R12 battery (U (voltage under load) ≈ 3.5 V; E (capacity) ≈ 0.6 A * h). At a voltage of 3.5 V, the stored energy of the battery is Q = 0.6 * 3.5 = 2.1 W * h, then r = 0.6 Ohm.

2) When the circuit is closed, the current will be: I = εi / (r + R) = 4.5 / (0.01 + 0.6) ≈ 7.4 A.

Answer: Since the internal resistance of the batteries fluctuates within a few ohms, it is impossible to get a current of hundreds of amperes.

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