Answer:
0.28 A
Explanation:
Given that the 30.0 ohm and 75.0 ohm resistor are connected in parallel, The equivalent resistance can be calculated by :
[tex] \: \dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}[/tex]
Plug in the values:
[tex]\dfrac{1}{R_{eq}} = \dfrac{1}{30} + \dfrac{1}{75}[/tex]
[tex]\dfrac{1}{R_{eq}} = \dfrac{5 + 2}{150} [/tex]
[tex] \dfrac{1}{R_{eq}} = \dfrac{7}{150} [/tex]
[tex] {R_{eq}} = \dfrac{150}{7} [/tex]
[tex]{ \boxed{ \pmb{{R_{eq}} = 21.42 \: \Omega}}}[/tex]
Calculating the current (I) :
The formula to calculate the current flowing out of the battery if the resistors are connected in parallel is :
[tex] \: \: \: \: \: I = \dfrac{V}{R} [/tex]
Where :
I is current in Amperes (A)
V is voltage in Volts (V)
R is Resistance in ohms [tex](\Omega)[/tex]
Plug in the values :
[tex]I = \dfrac{6}{21.42}[/tex]
[tex]{\boxed{ \pmb{ I =0.28 \: A}}}[/tex]
Hence, the the current flowing out of the battery if the resistors are connected in parallel is 0.28 A
