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- In this chapter, most examples and problems involved direct current (DC). DC circuits have the current flowing in one direction, from positive to negative. When the current was changing, it was changed linearly from I=ImaxtoI=+Imax and the voltage changed linearly from V=Vmax to V=+Vmax where Vmax=ImaxR .Suppose a voltage source is placed in series with a resistor of R = 10 that supplied a current that alternated as a sine wave, for example, I(t)=(3.00A)sin(24.00st) . (a) What would a graph of the voltage drop across the resistor V(t) versus time look like? (b) What would a plot of V(t) versus I(t) for one period look like? (Hint: If you are not sure, try plotting V(t) versus I(t) using a spreadsheet.)Item 10 Learning Goal: To understand the dynamics of a series R-C circuit. Consider a series circuit containing a resistor of resistance R and a capacitor of capacitance C connected to a source of EMF & with negligible internal resistance. The wires are also assumed to have zero resistance. Initially, the switch is open and the capacitor discharged. (Figure 1) Let us try to understand the processes that take place after the switch is closed. The charge of the capacitor, the current in the circuit, and, correspondingly, the voltages across the resistor and the capacitor will be changing. Note that at any moment in time during the life of our circuit, Kirchhoff's loop rule holds and, indeed, it is helpful: E-VR-Vc = 0, where VR is the voltage across the resistor and Vc is the voltage across the capacitor. Figure R www io=0 C = 90=0 1 of 1 Part F In the steady state, what is the charge q of the capacitor? Express your answer in terms of any or all of E, R, and C. ► View Available Hint(s)…Item 10 Learning Goal: To understand the dynamics of a series R-C circuit. Consider a series circuit containing a resistor of resistance R and a capacitor of capacitance C connected to a source of EMF & with negligible internal resistance. The wires are also assumed to have zero resistance. Initially, the switch is open and the capacitor discharged. (Figure 1) Let us try to understand the processes that take place after the switch is closed. The charge of the capacitor, the current in the circuit, and, correspondingly, the voltages across the resistor and the capacitor will be changing. Note that at any moment in time during the life of our circuit, Kirchhoff's loop rule holds and, indeed, it is helpful: E-VR-Vc = 0, where VR is the voltage across the resistor and Vc is the voltage across the capacitor. Figure & R ww io=0 C 90=0 1 of 1 Part B Immediately after the switch is closed, what is the voltage across the resistor? Submit Request Answer E Part C zero Immediately after the switch…
- Use the exact values you enter in previous answer(s) to make later calculation(s). Consider the following figure. (Assume R, = 32.0 0, R, = 21.0 0, andV = 18.0 V.) R 10.0 V 5.00 N R2 (a) Can the circuit shown above be reduced to a single resistor connected to the ba No. This multi-loop circuit does not contain any resistors in series (i e, connected so all the current in one must pass through the other) nor in parallel (connected so the voltage drop across one is always the same as that across the other). Thus, this circuit cannot be simplified any further, and Kirchhoff's rules must be used to analyze it. (b) Find the magnitude of the current and its direction in each resistor. R3: 21.0 N = X A 5.00 Q X A R: 32.0 N = X AОp Amp H.W) You have 5 voltage inputs (V1...5) and you wish to calculate 4(V1 + V2 + V3) – 2(V4 + V5). Design & Draw a circuit to achieve this.Consider an underdamped series LRC circuit with a variable resistor. What will happen to the charge oscillations of in the circuit if the resistance is slowly increased?
- A series circuit contains a resistor and a capacitor as shown below. Determine a differential equation for the charge q(t) on the capacitor if the resistance is R, the capacitance is C, and the impressed voltage is E(t). (Use E for E(t) and q for q(t).) dq dt E RExplain why the current is the same in all parts of a series circuit. Further, de- scribe why the IR drops are zero in an open series resistive circuit.#1. The capacitor in the figure is initially uncharged and the switch is at position c and not connected to either side of the circuit. At t = 0, the switch is flipped to position a for 20 ms,thenflipped back to position c for 10 ms, then flipped to position b for 20 ms, and finally flipped toposition c again. a) Using the Kirchhoff Voltage Law, write the differential equationsthat describethe circuit between t = 0 –20 ms andt = 30 –50 ms. b) Solve two differential equations you find ina) with appropriate initial condition to find the current through and the voltage across the capacitor as functions of time. c) Sketch the graphs of the current and voltage you find in b) from t = 0 to 60 ms. #2. Now the 40 uF capacitor in the circuit in #1 is replaced with a 0.4 H inductor. The inductor in this circuit is initially uncharged and the switch is at position c and not connected to either side of the circuit. At t = 0, the switch is flipped to position a for 20 ms, then flipped back to…
- Total resistance in a parallel circuit is less than your smallest resistor. The reason for this prob is that the total resistance has to divide, (or) split up. Can you help me to conceptualize this, which would get into the difference between parallel versus series circuit.If multiple capacitors are in a series circuit, describe how the total charge can be found. What if the circuit is a parallel circuit? Given a capacitance of 30 nF and a voltage of 30 J, compute for the charge and energy.The RC circuit shown on the right consists of a 1/µF capacitance, with an initial stored energy of 450 µJ and a lkN resistance. If the circuit is subjected to a dc forcing voltage Vs, and if it is found that v(2 ms) = 0 a) Find Vs b) How much voltage changes v(t = 0.1ms) to v(t = Ims)