Question

Explain how this proverb relates to the following circuit, where electric current from the batteryencounters two alternate paths, one being less resistive than the other:

Answer 1

The current always choose low resistor to flow in circuit

Explanation:

Gven:

V = 10 V

R1 = 250 Ω

R2 = 800 Ω

Apply Ohm's law:

`I=(V)/(R)`

We calculate the current in each case:

[tex]\begin{gathered} I_1=\frac{V}{R_1}=\frac{10}{250}=0.04\text{ A} \\ \\ I_2=\frac{V}{R_1}=\frac{10}{800}=0.0125\text{ A} \\ \\ \text{ Therefore:} \\ \\ I_2

That means current always choose low resistor to flow in circuit