Explain the advantages of using parallel circuits to series circuits in wiring rooms

1 answer:

5 0

The major advantage of parallel circuit which makes it advantageous in the application of wiring rooms are the supply of equal voltage across each circuit, which is not present in series circuit.

<u> Explanation:</u>

The use of parallel circuit is preferred over series circuit while wiring the room as the parallel gives the same voltage across every circuit which keeps it safe from damaging or burning as they give consistent voltage.

Apart from this, the parallel circuit allows you for independent operation without letting use of all the equipment present in the room. Then there is scope for additional components without changing the voltage.

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A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per m

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(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.

(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.

<h3>Electric field at a point on the Gaussian surface</h3>

The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;

E = λ/2πε₀r

where;

λ is linear charge density
ε₀ is permitivity of free space
r is the position of the charge

<h3>At a distance of 5.5 m</h3>

$E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C$

<h3>At a distance of 2.5 m</h3>

$E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C$

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.

Learn more about electric field here: brainly.com/question/14372859

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The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i

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To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

We can express this also as,

$log_{10}(10^6)$

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

$6*log_{10}(10)$