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A 9.0 volt battery is connected to four resistors in a parallel circuit. The resistors have 7.0 Ω, 5.0 Ω, 4.0 Ω, and 2.0 Ω respectively. What is the total current in the circuit?



Answer :

series,effective resistance =R₁+R₂+R₃...
parallel,effective resistance 1/R=1/R₁ +1/R₂ +1/R₃...
 
Here,effective resistance 1/R =1/7 +1/5+ 1/4+1/2  
                                           =1.092
                                       R = 1/1.092 =0.915Ω
voltage V=9 V 
current I=V/R
           I=9 / 0.915
            =9.83 A
parallel resistance = [tex] \frac{1}{7} [/tex] + [tex] \frac{1}{5} [/tex] + [tex] \frac{1}{4} [/tex] + [tex] \frac{1}{2} [/tex]
=[tex] \frac{140}{153} [/tex]

now , V= 9V
V = IR
I = [tex] \frac{V}{R} [/tex]
I = [tex] \frac{9}{140} [/tex] x 153
I = 9.835714 A

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