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3 years ago

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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 respe

ctively. What is the total current in the circuit?

2 answers:

lidiya [134]3 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]3 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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a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

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g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

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$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

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$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

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