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maria [59]
3 years ago
10

What is kirchoff s law???

Physics
2 answers:
kolbaska11 [484]3 years ago
8 0
There are two laws named for Kirchhoff.  The both concern electrical circuits.
Here they are in my own words:

1).  The sum of the voltage drops around any closed loop in a circuit is zero.

2).  The sum of the currents at any single point in a circuit is zero.
Luda [366]3 years ago
7 0
<em>there are two
<u>1 Kirchhoff's current law:
</u>it states that the sum of all currents flowing in all conductors that meet at single point is zero..
<u>2 Kirchhoff's voltage law:
</u>
it states that the algebraic sum of all voltages around any closed network is zero...

</em>
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Explanation:

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An example of pull as a force would be opening a door. ...

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3 years ago
True or false? An object at rest has an instantaneous acceleration of zero.
Tpy6a [65]
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2 years ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

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}}

                     = 2.42 \times 10^{-2} C/m^{3}

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}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(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

7 0
3 years ago
An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux
zhenek [66]

Answer:

\phi=1.56\times 10^{-5}\ Wb

Explanation:

Given that,

Emf, V = 22 mV

Number of turns in the coil us 519

Rate of change of current is 10 A/s.

We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.

Let's find the inductance first. So,

L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H

We have,

L=\dfrac{N\phi}{I}, \phi is magnetic flux

\phi=\dfrac{LI}{N}\\\\\phi=\dfrac{0.0022\times3.7}{519}\\\\\phi=1.56\times 10^{-5}\ Wb

So, the magnetic flux is 1.56\times 10^{-5}\ Wb.

8 0
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AveGali [126]

Answer:

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