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4 years ago

9

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 respe

ctively. What is the total current in the circuit?

2 answers:

lidiya [134]4 years ago

5 0

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

shepuryov [24]4 years ago

4 0

Parallel resistance = <u /> $\frac{1}{7}$ + $\frac{1}{5}$ + $\frac{1}{4}$ + $\frac{1}{2}$
= $\frac{140}{153}$

now , V= 9V
V = IR
I = $\frac{V}{R}$
I = $\frac{9}{140}$ x 153
I = 9.835714 A

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Answer:

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Leg 2 -> 22.44 m/s

Leg 3 -> 11.20 m/s

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Time, t = 2.85 min = 171 s

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Initial velocity, u = 22.44 m/s

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Substituting

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b) Leg 1:

$v_{avg}=\frac{123.42}{11}=11.22m/s$

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Displacement is the shortest distance between two points. Here, displacement between the starting and end points is the vector $\overrightarrow{OA}$ .

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