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2 years ago

Find the voltage across the 15 Q resistor. [?] V No links please

Physics

1 answer:

Margaret [11]2 years ago

4 0

Answer:

Explanation:

same idea as before Liam, first, find the parallel resistance in 35 || 20

(35*20) / (35+20) = 700 / 55 = 12.727272 ohms

now add the 12.727272 + 15 = 27.727272 ohms total resistance

V = IR

10 = I * 27.727272

10 / 27.727272 = I

0.360655 = I

V = IR (again, but across the 15 ohm resistor)

V = 0.360655 * 15

V = 5.4098

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A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre

disa [49]

Answer:

I = 26.36 cosω t A

Explanation:

Given that

C=0.74 mF

Vrms= 82 V

Frequency ,f= 49 Hz

We know that ω = 2 π f

ω = 2 x π x 49

ω = 307.72 rad/s

As we know that voltage given as

V= Vo sinω t

$V_o=\sqrt2\ V_{rms}$

$V_o=\sqrt2\ \times 82$ \

Vo=115.96 V

V=115.96 sinω t

The current given as

$I=C\dfrac{dV}{dt}$

$I=0.74\times \dfrac{dV}{dt}\ mA$

$\dfrac{dV}{dt}=115.96\omega cos\omega t$

$I=0.74\times 115.96\times 307.22 cos\omega t\ mA$

I = 26362.67 cosω t mA

I = 26.36 cosω t A

This is the current at time ant time t.

7 0

3 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

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3 years ago

Which of the following statements is true? A. X-rays have longer wavelengths than microwaves. B. Radio waves have shorter wavele

3241004551 [841]

D is correct, Gamma rays have the shortest wavelength.

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3 years ago

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Artemon [7]

Explanation:

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3 years ago

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avanturin [10]

Answer:

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Explanation:

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3 years ago