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A circuit contains four capacitors in parallel (10 F, 3 F, 7 F, and 1 F). What is the equivalent capacitance of this circuit?

Oxana [17]

The equivalent capacitance ( $C_{eq}$ ) of an electrical circuit containing four capacitors which are connected in parallel is equal to: A. 21 F.

<h3>The types of circuit.</h3>

Basically, the components of an electrical circuit can be connected or arranged in two forms and these are;

Series circuit

Parallel circuit

<h3>What is a parallel circuit?</h3>

A parallel circuit can be defined as an electrical circuit with the same potential difference (voltage) across its terminals. This ultimately implies that, the equivalent capacitance ( $C_{eq}$ ) of two (2) capacitors which are connected in parallel is equal to the sum of the individual (each) capacitances.

Mathematically, the equivalent capacitance ( $C_{eq}$ ) of an electrical circuit containing four capacitors which are connected in parallel is given by this formula:

Ceq = C₁ + C₂ + C₃ + C₄

Substituting the given parameters into the formula, we have;

Ceq = 10 F + 3 F + 7 F + 1 F

Equivalent capacitance, Ceq = 21 F.

Read more equivalent capacitance here: brainly.com/question/27548736

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3 0

2 years ago

Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe

Greeley [361]

Answer:cho v₀ =0s

α=Δv/Δt

Explanation:

$\frac{0-495}{2,16-1,78}$

=-1302,631579

chuyển động chậm dầnđều

3 0

3 years ago

A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito

Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

$c = \frac{c1 \times c2}{c1 + c2}$

$= \frac{2 \times 4}{2 + 4}$

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

$= 1.33 \times 10 {}^{ - 6} \times 8$

$= 10.64 \times 10 {}^{ - 6}$

charge on 2.0μf capacitor is

$\frac{Qeq}{2 \times 10 {}^{ - 6} }$

$= \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }$

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

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3 0

2 years ago

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

4 0

3 years ago

Secondary evidence is the basis for drawing scientific conclusions is this true ir false

Leto [7]

The statement: "secondary evidence is the basis for drawing scientific conclusions" is definitely false. Secondary evidence is the body of information obtained to prove the existence of unknown or missing primary evidence. In drawing scientific conclusions, primary evidence is needed.

7 0

3 years ago

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