Question

Three resistors, 3.0 Ω, 12 Ω, and 4.0 Ω, are connected in parallel across a 6.0-V battery as shown above. What is the current through the battery?

Answer 1

the current flowing through the battery is 4 A.

Step-by-step explanation:

Given;

resistance of the three resistors in parallel, R₁, R₂ and R₃ = 3.0 Ω, 12 Ω, and 4.0 Ω

voltage of the battery, V = 6.0 V

The equivalent resistance is calculated as follows;

`(1)/(R_T) = (1)/(R_1) + (1)/(R_2) + (1)/(R_3) \\\\(1)/(R_T) = (1)/(3) + (1)/(12) + (1)/(4) \\\\(1)/(R_T) = (4 \ + \ 1 \ + \ 3)/(12) \\\\(1)/(R_T) = (8)/(12) \\\\(1)/(R_T) = (2)/(3) \\\\R_T = (3)/(2) \\\\R_T = 1.5 \ ohms`

The current flowing through the battery is calculated as follows;

`I = (V)/(R_T) \\\\I = (6)/(1.5) \\\\I = 4 \ A`

Therefore, the current flowing through the battery is 4 A.