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?
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 = \(\frac{1}{7}\) + \(\frac{1}{5}\) + \(\frac{1}{4}\) + \(\frac{1}{2}\)
=\(\frac{140}{153}\)
now , V= 9V
V = IR
I = \(\frac{V}{R}\)
I = \(\frac{9}{140}\) x 153
I = 9.835714 A