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2300_Chapter6_Sample_probs

# 2300_Chapter6_Sample_probs - The circuit shown below has a...

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The circuit shown below has a switch which closed at t = 0. The voltages v 1 and v 2 were measured before the switch was closed, and it was found that 1 ( ) 15[V], for 0, and v t t = < 2 ( ) 7[V], for 0. v t t = - < In addition, for time greater than zero, it was determined that ( 29 [ ] 1 500 s ( ) 22 mA , for 0. t R i t e t - - = Explore the energy stored in the capacitors for t < 0, and for t = . 1[k ] + - v 1 C 1 = 6[ μ F] C 2 = 3[ μ F] t = 0 + - v 2 i R (t) Solution: For t 0, we have no current through the capacitors, and therefore no change in the voltages across the capacitors. Thus, we can write ( 29 ( 29 ( 29 2 , 1 1 1 2 6 , 1 1 (0) (0) , or 2 1 (0) 6 10 [F] 15[V] 675[ J]. 2 STO C STO C w C v w μ - = = × = ( 29 ( 29 ( 29 2 , 2 2 2 2 6 , 2 1 (0) (0) , or 2 1 (0) 3 10 [F] 7[V] 73.5[ J]. 2 STO C STO C w C v w μ - = = × - = So, the total stored energy, in the combination of the two capacitors, is 748.5[ μ J]. Now, once the switch closes, current begins to flow. This makes sense, since when the switch closes, we now have 22[V] across the resistor, so current must flow. If you think about it, it makes sense that the current will decrease with time. As the charges move, the voltages across the capacitors will increase and decrease, so that the voltage across the resistor will decrease. When the current decreases to zero, the two voltages v 1 and v 2 will be equal. What value will these voltages have? We can solve this by integrating the current i R (t) , with the defining equation for the capacitor, that is

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2 2 2 0 1 1 1 0 1 ( ) ( ) (0), and 1 ( ) ( ) (0). R R v i s ds v C v i s ds v C ∞ = + - ∞ = + However, there is a quicker way to solve this circuit. Note that charge must be conserved. Therefore, we know that the total charge on the two capacitors at the beginning, must be equal to the total charge on the two capacitors at the end. So, we can write , 1 1 2 2 , 1 1 2 2 (0) (0) ( ) ( ). TOTAL BEFORE TOTAL AFTER q C v C v q C v C v = + = = ∞ + Of course, as we argued earlier, the two final values must be equal. That is, 1 2 ( ) ( ) . FINAL v v v ∞ = ∞ = So, we plug this into our equation, and we get ( 29 ( 29 1 1 2 2 1 2 1 1 2 2 1 2 (0) (0) , or 6 15[V] 3 7[V] (0) (0) 7.667[V]. 9 FINAL FINAL FINAL C v C v C v C v C v C v v C C + = + + - + = = = + You can check the integrals above to make sure this is correct. Now, let’s look at the energy after a long time. We get ( 29 ( 29 ( 29 2 , 1 1 2 6 , 1 1 ( ) , or 2 1 ( ) 6 10 [F] 7.667[V] 176.3[ J].
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2300_Chapter6_Sample_probs - The circuit shown below has a...

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