Answer a)0.987 b)6.9x10^-4J c)8.00x10^-4 d)5320 Ohms A 0.01 Mu F capacitor, a 0.
ID: 1507481 • Letter: A
Question
Answer
a)0.987
b)6.9x10^-4J
c)8.00x10^-4
d)5320 Ohms
A 0.01 Mu F capacitor, a 0.1 H inductor whose resistance is 1000 Ohms, and switch are connected in a series circuit. The capacitor is initially is charged to a potential difference of 400 V. The switch is then closed. Find the ratio of the oscillation frequency of the actual circuit the frequently it would have if the resistance were zero. How much energy is converted to heat in the first complete cycle? How much energy is converted to heat in the first complete train of oscillations? How much additional resistance can be put into the circuit before the circuit becomes non-oscillatory?Explanation / Answer
Hi,
As we have an RLC circuit in series whose resistance is small we can apply the following equation:
Q(t) = Qmax exp[ -(R/2L)t ] cos(t) ; where Qmax is the charge stored in the capacitor, R is the resistance, L is the inductance and is the oscillation frequency, which in turn can be calculated as:
= [ 1/LC - (R/2L)2 ]1/2
(a) The value of the oscillation frequency of the actural circuit is:
= [ 1/(0.1 H*0.01*10-6 F) - (1000 /2*0.1 H)2 ]1/2 = 31224.99
The value of the oscillation frequency if the resistance were equal to cero is:
= [ 1/(0.1 H*0.01*10-6 F) ]1/2 = 31622.78
The fraction they ask is then equal to:
f = (31224.99 / 31622.78) = 0.987
(d) The critical resistance that the circuit can handle and still have oscillations is:
Rc = (4L/C)1/2 = (4*0.1 H/0.01*10-6 F)1/2 = 6325
So the max amount of resistance that can be added to the circuit is:
R' = Rc - R = 6325 - 1000 = 5325
(b) The energy converted to heat by the resistance can be found by:
E = QV ; where Q is the charge and V is the initial voltage. In other terms:
E = (1 - f) CV2 ; where f is the fraction of the maximun charge.
If it has passed a cycle, then:
t = 2/ = 2/(31224.99) = 2.01*10-4
Q(t) = Qmax exp( -5000*2.01*10-4) = 0.366 Qmax :::::::: f = 0.366
E = (1- 0.366)*CV2 = 0.634*(0.01*10-6 F)(400 V)2 = 1.01*10-3 J
(c) Now, if it has passed the complete train of oscillations it means that the time is very long but it fulfills the next property:
t = n ( 2/ ) ; where n is a positive integer number which must be really big.
If t is like that, then we have the following:
cos(t) = 1 and exp[-(R/2L)t] = 0 therefore f = 0 and;
E = CV2 = (0.01*10-6 F)(400 V)2 = 1.6*10-3 J ; this means that all the energy stored in the circuit was turned into heat.
Note: I'm concious about the difference in the results of parts (b) and (c) , but I think that can be because we assumed that the resistance of the circuit was small, if that is not true, then other equations are in order.
I hope it helps.
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