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never [62]
3 years ago
12

If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?

Physics
2 answers:
MAXImum [283]3 years ago
8 0

Answer: 1.2 amps

Explanation:

i got it right in my exam :D

Shalnov [3]3 years ago
6 0

In a series circuit the total current is the same throughout resistors and so:

I_{total}=I_1=I_2

The voltage is distributed throughout the resistors and so:

V_{total}=V_1+V_2

and the total resistance can be calculated by adding up the resistors resistance:

R_{total}=R_1+R_2

First thing is to calculate the total resistance and so:

R_{total}=6\Omega + 4\Omega = 10\Omega

And by Omh's law V=IR we have:

V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.


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3 years ago
Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m
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18 N/C

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5 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

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4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

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The electric field between plates, E ≈ 1.1294 × 10¹² N/C

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The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

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The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

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