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3 years ago

Which circuit is a series circuit ?? plz help

Physics

2 answers:

gregori [183]3 years ago

7 0

In a series circuit, all of the components are connected in the same 'loop' and the current only has one direction/path it can flow through.

In the first three options, the current has multiple paths it can go through. So these three circuits are parallel and not series.

In the last option, the current only has one path where it can flow through, so that circuit is in series.

So Circuit <u>D </u>is a series circuit.

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Answer

Circuit D

Ugo [173]3 years ago

5 0

Answer:the last one

Explanation:in a series circuit, only one current flows through it. i.e it connected by a single wire

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3 years ago

Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 < x a. 283 eV <br> b. 339 eV <br> c.

denis23 [38]

This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113 eV

d. 226 eV

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

$E_{nx,ny,nz$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( $n_x = n_y = n_z = 1$ )

$( n_x^2 + n_y^2 + n_z^2 )$ × π²h"² / 2ml²

since h" = h/2π

$( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

so we substitute

$E_{111$ = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

$E_{111$ = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]

$E_{111$ = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

$E_{111$ = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

$E_{111$ = 3 × [ 37.645578 ]

$E_{111$ = 112.9 ≈ 113 eV

$E_{211$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

we substitute

$E_{211$ = ( 1² + 1² + 2² ) × [ 37.645578 ]

$E_{211$ = 6 × [ 37.645578 ]

$E_{211$ = 225.87 ≈ 226 eV

$E_{221$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

we substitute

$E_{221$ = ( 2² + 2² + 1² ) × [ 37.645578 ]

$E_{211$ = 9 × [ 37.645578 ]

$E_{211$ = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

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2 years ago