For the circuit in Fig. 8.58, the capacitor voltage at t = 0 − (just before the switch is closed) is: (a) 0 V (b) 4 V (c) 8 V (d) 12 V Figure 8.58 For Review Questions 8.1 and 8.2.
For the circuit in Fig. 8.58, the capacitor voltage at t = 0 − (just before the switch is closed) is: (a) 0 V (b) 4 V (c) 8 V (d) 12 V Figure 8.58 For Review Questions 8.1 and 8.2.
For the circuit in Fig. 8.58, the capacitor voltage at t = 0− (just before the switch is closed) is:
(a) 0 V
(b) 4 V
(c) 8 V
(d) 12 V
Figure 8.58
For Review Questions 8.1 and 8.2.
Expert Solution & Answer
To determine
Choose the correct option to find the capacitor voltage at t=0− (just before the switch is closed).
Answer to Problem 1RQ
The correct option form the given choices is (a) 0 V.
Explanation of Solution
Calculation:
Redraw the given circuit as shown in Figure 1.
For the DC circuit, at the steady state condition when switch is open at t=0−, the capacitor acts as open circuit and the inductor act as short circuit. Therefore, the Figure 1 becomes as shown in Figure 2.
Since, the capacitor is open circuited, the voltage across the capacitor at t=0− is 0 V.
Therefore, the capacitor voltage at t=0− is vC(0−).
vC(0−)=0V
Therefore, the option (a) is correct and the options (b), (c), (d) are incorrect.
Conclusion:
Thus, the correct option from the given choices is (a) 0 V.
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Q8. What is the charge of the second capacitor(C2)?
A) 64mc
B) 32mc
C) 16me
D) 48mc
E) 24mc
C2=8mf]
C1=6mf
C3-4 mf
V=12V
GENERAL RLC CIRCUITS:
Solve and show your solution completely. Make sure your answers are in 4 decimals places.
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effectivecurrent of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
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