Answer:
[tex]Branch Currents: \\I_{R1} = 0.2 A\\I_{R2} = 0.4 A\\I_{R3} = 0.6 A\\\\Equivalent Resistance [Total Resistance(R_{T} )]:\\R_{T} = 5ohms\\\\Current through the battery [Total Current(I_{T} )]:\\I_{T} =1.2A[/tex]
Explanation:
Given:
[tex]R_{1} = 30ohms \\R_{2} = 15ohms \\R_{3} = 10ohms\\V= 6V[/tex]
Using Ohm's Law:
[tex]V=IR\\[/tex] ⇒ [tex]I=\frac{V}{R}[/tex]
To get the total resistance of a parallel circuit:
[tex]R = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}} + ... + \frac{1}{R_{n} } }[/tex]
[tex]R = \frac{1}{\frac{1}{30} +\frac{1}{15} +\frac{1}{10} } = 5ohms[/tex]
To get the branch current by applying Ohm's Law:
[tex]I_{R1} = \frac{6volts}{30ohms} = 0.2 A[/tex]
[tex]I_{R2} = \frac{6volts}{15ohms} = 0.4 A\\[/tex]
[tex]I_{R3}= \frac{6volts}{10ohms} = 0.6 A[/tex]
To get the total current:
[tex]I_{T} = \frac{V_{T} }{R_{T} } \\I_{T} = \frac{6volts}{5ohms} \\I_{T} = 1.2 A[/tex]
(to check if the current is correct you may add all the branch currents)
[tex]I_{T} = I_{1} + I_{2} + I_{3} \\I_{T} = 0.2 A + 0.4 A +0.6A\\I_{T} = 1.2 A[/tex]
