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Mazyrski [523]
3 years ago
8

three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,

and the charge on each condenser is 6×10^-8(a) calculate the capacitance of each condenser (b)calculate the effective capacitance of the combination
Physics
1 answer:
ioda3 years ago
5 0

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF

Capacitance of capacitor 2,

C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF

Capacitance of capacitor 3,

C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF

(b) The equivalent capacitance in series combination is :

\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF

Hence, this is the required solution.

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Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

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the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

6 0
3 years ago
If a current is two amps and the resistance is 3 ohms, how much voltage was needed?
Hunter-Best [27]

Answer:

6 V

Explanation:

We can solve the problem by using Ohm's law:

V=RI

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V is the voltage in the circuit

R is the resistance

I is the current

In this problem, we know the current, I=2 A, and the resistance, R=3 \Omega, therefore we can find the voltage in the circuit:

V=RI=(3 \Omega )(2 A)=6 V

7 0
3 years ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

7 0
2 years ago
A 20 metric ton train moves toward the south at 50 m/s. At what speed must it travel to have four times its original momentum
OverLord2011 [107]

Answer:

200 m/s

Explanation:

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5 0
2 years ago
A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
3 years ago
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