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3 years ago

A 12 A fuse is placed in a parallel circuit that has two branches. 8 A flows in branch 1 and 6 A flows in branch 2. This fuse

Physics

1 answer:

kaheart [24]3 years ago

3 0

Answer:

1. will blow because the total current in this circuit is 14 A which is greater than 12 A.

Explanation:

According to Kirchoff current law (KCL) which states that the total current flowing in a circuit is equal to the sum of the individual branch current.

If the supply current is greater than the sum of the individual branch current, then the load will collapse or blow off.

In the question given, the total current of the fuse is 12A

Sum of branch currents = current in branch 1 + current in branch 2

= 8A+6A

= 14A

As we can see that the supply current is lower than the sum of the branch current, this will cause the fuse to blow because some of the branch current will be sent back on the fuse and thereby causing the fuse to blow.

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Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

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Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

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(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

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