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bixtya [17]
2 years ago
12

Explain Resistor in parallel and series. ​

Physics
2 answers:
patriot [66]2 years ago
6 0

\sf\huge\underline\blue{Resistor:-}

\rightarrow<u>A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element</u>.

\rightarrowResistors reduce the current flow and lower voltage levels within circuits.

\sf\large\underline\purple{Resistors \:in\: Series:-}

\rightarrowA <u>circuit is said to be connected in series</u> when the same amount of <u>current flows through the resistors</u>. In such circuits, the voltage across each resistor is different.

\rightarrowIn a series connection, if any resistor is broken or a fault occurs, then the entire circuit is turned off. The construction of a series circuit is simpler compared to a parallel circuit.

\rightarrowFor the above circuit(attached image-1), the total resistance is given as:

\sf{R_{total}\: = \:R1 + R2 + ….. + Rn}

The total resistance of the system is just the total of individual resistances.

\sf\large\underline\purple{Resistors \:in\: Parallel:-}

\rightarrowA <u>circuit is said to be connected in parallel</u> when the <u>voltage is the same across the resistors</u>. In such circuits, the current is branched out and recombines when branches meet at a common point.

\rightarrowA resistor or any other component can be connected or disconnected easily without affecting other elements in a parallel circuit.

\rightarrowThe figure(attached image -2) above shows ‘n’ number of resistors connected in parallel. The following relation gives the total resistance here

\sf{\frac{1}{R_{total}}\: = \:\frac{1}{R1} + \frac{1}{R2} + ….. + \frac{1}{Rn}}

\rightarrowThe sum of reciprocals of resistance of an individual resistor is the total reciprocal resistance of the system.

_______________________________

Hope it helps you:)

Strike441 [17]2 years ago
4 0

Answer:

In a series circuit, the output current of the first resistor flows into the input of the second resistor; therefore, the current is the same in each resistor. In a parallel circuit, all of the resistor leads on one side of the resistors are connected together and all the leads on the other side are connected together.

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Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.
Dennis_Churaev [7]

Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

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3 years ago
As your rockets went upwards how would you describe how the energies changed?
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If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo
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Read 2 more answers
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Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

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q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

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a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

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T = 2Πr / v

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T = 1.19*10⁻⁷s

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K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

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