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tankabanditka [31]

3 years ago

7

You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2

. Assume L2 > L1. If the resistors are connected in series with a battery, which resistor has the most power delivered to it? Explain. b: If they are connected in parallel with a battery, which resistor has the most power delivered to it? Explain.

1 answer:

oee [108]3 years ago

8 0

Answer:

Explanation:

Given

Resistor A has length $L_1$

and Resistor B has Length $L_2$

and Resistance is given by

$R=\frac{\rho L}{A}$

Considering $\rho$ and A to be constant thus

$R_2>R_1$ because $L_2>L_1$

(a)When they are connected in series

As the current in series is same and power is $i^2R$

therefore $P_2>P_1$ as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

$P=\frac{V^2}{R}$

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

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dolphi86 [110]

Answer:

0.785 m/s

Explanation:

Hi!

To solve this problem we will use the equation of motion of the harmonic oscillator, <em>i.e.</em>

$x(t) = A cos(\omega t )+B sin(\omega t)$ - (1)

$\frac{dx}{dt}(t) = \omega (B cos(\omega t )- A sin(\omega t)$ - (1)

The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:

$x(0) = 0.100$

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Since cos(0)=1 and sin(0) = 0:

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$\frac{dx}{dt}(0) = -B\omega$

We get

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Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:

$x(0.4) = - 0.100$

Since

$x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)$

This is the same as:

$-1 = cos(0.4\omega)$

We know that cosine equals to -1 when its argument is equal to:

(2n+1)π

With n an integer

The first time should happen when n=0

Therefore:

π = 0.4ω

or

ω = π/0.4 -- (2)

Now, the maximum speed will be reached when the potential energy is zero, <em>i.e. </em>when the sping is not stretched, that is when x = 0

With this info we will know at what time it happens:

$0 = x(t) = 0.100cos(\omega t)$

The first time that the cosine is equal to zero is when its argument is equal to π/2

<em>i.e.</em>

$t_{maxV}=\pi /(2\omega)$

And the velocity at that time is:

$\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\$

But sin(π/2) = 1.

Therefore, using eq(2):

$\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4$

And so:

$V_{max} = \pi / 4 =0.785$

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