Answer:
the current flowing through the battery is 4 A.
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;
[tex]\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \\\\\frac{1}{R_T} = \frac{1}{3} + \frac{1}{12} + \frac{1}{4} \\\\\frac{1}{R_T} = \frac{4 \ + \ 1 \ + \ 3}{12} \\\\\frac{1}{R_T} = \frac{8}{12} \\\\\frac{1}{R_T} = \frac{2}{3} \\\\R_T = \frac{3}{2} \\\\R_T = 1.5 \ ohms[/tex]
The current flowing through the battery is calculated as follows;
[tex]I = \frac{V}{R_T} \\\\I = \frac{6}{1.5} \\\\I = 4 \ A[/tex]
Therefore, the current flowing through the battery is 4 A.
