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2 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]2 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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2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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3 years ago

Name three ways that being obese puts a strain on your heart & can lead to serious health problems.

Wittaler [7]

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2 years ago

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A 60-watt light bulb has a voltage of 120 volts applied across it and a current of 0.5 amperes flows through the bulb. What is t

Naddik [55]

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V=120volts

From ohms law

V=IR

R=V/I

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R= 240ohms

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3 years ago