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3 years ago

What would the overall resistance be in a series circuit if there's an unknown R, 3Ω,3Ω and a current of 2A?

Physics

1 answer:

irinina [24]3 years ago

3 0

Answer:

R + 6 ohms

Explanation:

The current is the same in all three resistors

i = 2 amps

The problem is that the total voltage is unknown.

R = r

R1 =3 ohms

R2 = 3 ohms

The voltage drop is E = I*R

I = 2 amps

R1 = 3 ohms

E = 3*2 = 6 volts

The second 3 ohm resistor also has 6 volts across it.

The total is 12 volts so far. But we know nothing more on how to solve for R.

All you can say is that the total resistance = R + 6 ohms

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago

Use Hooke's Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length vari

Phantasy [73]

Answer:

706.68 N

Explanation:

By Hooke's law,

$F = ke$

$k=\dfrac{F}{e}$

Using the values in the question,

$k=\dfrac{265\text{ N}}{0.15 \text{ m}}=1766.7\text{ N/m}$

When e = 0.4 m,

$F = 1766.7\text{ N/m}\times0.4\text{ m}=706.68\text{ N}$

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3 years ago

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Answer:

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deff fn [24]

Answer:

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Explanation:

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3 years ago