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Georgia [21]
3 years ago
8

Which statement best describes the movement of electrical current when a capacitor is used in a circuit?

Physics
1 answer:
Masteriza [31]3 years ago
4 0

Answer: B.

Capacitors prevent current from moving through a circuit 

Explanation:

If a direct voltage is applied on the capacitor, no conduction current flows through the capacitor if its insulating medium is perfect insulator. This is due to the fact that there are no free charge carriers in such medium. Basically the real insulator contains very few charge carriers and therefore a very small leakage current passes in the capacitor depending on the conductivity of the insulator.

If an alternating voltage is applied on the capacitor, a displacement current passes through the capacitor irrespective of the insulating medium. This current is termed also the capacitive current. It flows because of changing electric displacement with time.

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A person walks 20.0° north of east for 3.20 km. How far would she have to walk due north and due east to arrive at the same loca
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Answer:

1.09 km, 3 km

Explanation:

displacement, d = 3.20 km at 20° North of east

A vector quantity has two components one along the x axis and the other is along Y axis. the component along X axis is called the horizontal component and it is due east.

The component along y axis is called vertical component and it is due North.

Displacement due North, dy = 3.20 Sin 20° = 1.09 km

Displacement due east, dx = 3.20 Cos 20° = 3 km

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4 years ago
Let us be two cylindrical conductors connected in parallel, to which a potential difference of V = 170V is applied. The two cond
PSYCHO15rus [73]

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

<h3>Resistance</h3>

Resistance is a measure of the opposition to flow of electric current. It is measured in ohms.

It is given by the formula:

R=\rho\frac{l}{A} \\\\where\ l=length,A=area,\rho=resistivity

Given that R₂ = 469 ohm, hence:

R_2=\rho\frac{l_2}{A_2} \\\\469=\rho\frac{l_2}{\pi r_2^2}

But l₁ = 6l₂, r₁ = (1/5)r₂, hence:

R_1=\rho \frac{l_1}{A_1}=\rho *\frac{6l_2}{[\pi (1/5)r_2]^2} =150 * \rho \frac{l_2}{[\pi r_2]^2}=30*469=70350\ ohm

The equivalent resistance (R) is:

R=\frac{R_1R_2}{R_1+R_1}=\frac{469*70350}{469+70350}  =466\ ohm

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

Find out more on resistance at: brainly.com/question/17563681

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What is background radiation?
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The uniform microwave radiation remaining from the Big Bang 
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