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3 years ago

If you add a battery to a series circuit what will happen to the current in the circuit

1 answer:

4 0

If there wasn't any battery before, then there was no current
in the circuit before, and there IS one now. That's just about
the greatest change possible.

If there WAS a battery there before and you added another one
in series with it, then there are a few different possibilities for the
effect on the current in the circuit:

-- If the new battery has the same voltage as the original one,
AND you connect the new one so that they're both in the same
direction, then the current in the circuit will become double the
original current (twice as much as it was before).

-- If the new battery has the same voltage as the original one, AND
you connect the new one so that they're in opposite directions, then
the two batteries cancel each other, the total voltage becomes zero,
and the current in the circuit completely disappears.

-- If the voltage of the two batteries is different AND you connect
the new one so that they're both in the same direction, then the
current in the circuit increases, by a factor of

   (sum of the two battery voltages)
divided by
   (voltage of the original battery alone).

-- If the voltage of the two batteries is different AND you
connect the new one so that they're in opposite directions,
then the current in the circuit decreases, by a factor of

   (difference of the two battery voltages)
divided by
      (voltage of the original battery alone)

and the current flows in the direction of whichever battery has
the greater voltage. If the new battery has greater voltage than
the original one alone, then the current reverses, and flows in
the opposite direction.

I think that covers all the possibilities.

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When sedimentary rock is exposed to heat and pressure what does it change into

kherson [118]

When sedimentary rock is exposed to heat and pressure it changes into METAMORPHIC ROCK.
Heat and pressure have the capacity to change a rock into a completely new type of rock. An igneous rock or a sedimentary rock can be changed into a metamorphic rock as a result of heat and pressure which the rock is subjected to. Metamorphic rocks are usually formed from already existing rocks that are exposed to pressure and heat.

8 0

3 years ago

Read 2 more answers

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display

goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .

This can be written as $\lambda=\frac{h}{p}$ .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

$p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}$

(b) As we know that kinetic energy is given by

$K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p$ ...(3)

Where mass of electron is $9.1\times10^-^3^1 kg$ .

Putting K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}$

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}$

(d) Putting E = 0.3eV in equation (2) , we get

$p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}$

(e) Putting $p=3eV/c$ in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}$

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

Learn more about de brogile here :

brainly.com/question/28165547

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6 0

2 years ago

Which object forms when a supergiant explodes? a red giant a black hole a white dwarf a neutron star

Blababa [14]

Answer:

a neutron star

Explanation:

6 0

3 years ago

Read 2 more answers

Which of the following is an acidic pH? <br> A. 6.0 <br> B. 7.0 <br> C. 8.0 <br> D. 9.0

lubasha [3.4K]

The answer is a 6.0 anything above 7 is alkaline

3 0

3 years ago

Read 2 more answers

Forţa de rezistenţă la înaintare pe care o întâmpină o maşină reprezintă o fracţiune f=0,15 din greutatea ei. Ce putere are moto

frosja888 [35]

Answer:

777.59N

Explanation:

(The forward strength of a car is a fraction f = 0.15 of its weight. What power does the engine have if the mass of the car is 1.5t and reaches a top speed of 144km / h?)

F = Mass * acceleration

Acceleration =change in velocity / time

F = Mv / t

F = 0.15mg

M = 1.5t

g = 9.8 m/s

V = 144 km/hr= (144 * 3600) / 1000 = 518.4 m/s

0.15mg = mv / t

0.15 * g = v/t

0.15 * 9.8 = 518.4 / t

1.47t = 518.4

t = 518.4 / 1.47 = 352.65s

But mass = 1.5t

Mass = 1.5 * 352.65 = 528.97kg

F = mv / t

F = (528.97 * 518.4) / 352.65

F = 777.59N

5 0

4 years ago