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3 years ago

if a current of 0.01 A is flowing through a circuit with two resistors in a series what is the voltage?

Physics

1 answer:

NikAS [45]3 years ago

8 0

Answer:

V = 0.01(R₁ + R₂)

Explanation:

From Ohm's law, the voltage V across a resistor with resistance R and a current I passing through it is given by

V = IR

If we have two resistances, R₁ and R₂ in series, the total resistance R is the sum of their resistances.

So, R = R₁ + R₂

Now, if the current in the circuit is 0.01 A, the voltage V is

V = IR

V = 0.01R

Substituting R = R₁ + R₂ into the equation for V, we have

V = 0.01(R₁ + R₂)

So, the voltage in two series resistors with current 0.01 A is

V = 0.01(R₁ + R₂)

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3 years ago

Read 2 more answers

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mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

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3 years ago

A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D

LenaWriter [7]

Answer:

vf = √(vi²+2*(F/m)*D)

Explanation:

Given

Mass of the particle: M

Initial speed of the particle: vi

Force: F

Distance: D

We can apply the formula

F = M*a ⇒ a = F/m

then we use the equation

vf = √(vi²+2*a*D)

⇒ vf = √(vi²+2*(F/m)*D)

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3 years ago