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myrzilka [38]
3 years ago
6

Electricity seems like a fundamental element in the construction of any home. However, to have electricity run into and out of a

home, there must be a complete circuit. Write a brief essay describing the minimum requirements for any electric circuit. Within your essay, describe the task for each of these elements using your lessons and experience within this virtual lab. 150 words max
Physics
1 answer:
olganol [36]3 years ago
8 0

Answer:

•Most of the components that make up curcuits are conductors. Electricity flows easily thought conductors such as metal or water.Electricity does not flow easily thought insulators such as wood or rubber.

•When a battery is used to power a circuit, the circuit the current must travel from one pole of the battery to the opposite battery.

•Resisotors show the flow of Electricity throught a circuit this can prevent components of the circuit from 'brning out'. In a curcuit diagram, resistors are represented by bending lines.

Explanation:

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Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
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Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

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Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

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