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Alik [6]
2 years ago
12

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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disa [49]

Answer:

I =  26.36 cosω t A

Explanation:

Given that

C=0.74 mF

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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
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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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