Question

Батарейка для карманного фонаря с Э,Д,С, 4,5 В при замыкании на сопротивление 7,5 Ом создает в цепи силу тока 0,5 А. Определить силу тока при коротком замыкании

Answer 1

Calculation of Current in the Circuit

To determine the current in the circuit when the pocket flashlight battery with an electromotive force (EMF) of 4.5 V is short-circuited with a resistance of 7.5 Ω, we can use Ohm's Law.

Ohm's Law states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). Mathematically, it can be represented as:

I = V / R

Given: - EMF (V) = 4.5 V - Resistance (R) = 7.5 Ω

Substituting the values into the formula, we can calculate the current:

I = 4.5 V / 7.5 Ω = 0.6 A

Therefore, the current in the circuit when the battery is short-circuited is 0.6 A.

Please note that the calculated value of 0.6 A is an approximation based on the given information. The actual value may vary depending on the characteristics of the battery and the circuit.