Problem 4. (4 points) LC circuits. In the circuit of Fig. 3, the battery emf is

Question

   Problem 4. (4 points) LC circuits. In the circuit of Fig. 3, the battery emf is 50 V, the resistance is 125 , and the capacitance is 4.5 F (it’s microfarads). The switch S is closed for a long time interval, and zero potential difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum value of 50 V. What is the value of the inductance?
In the problem below, the given resistance is 125 ohms and the circuit has 2 resistors. So, when calculating the current, would you divide 50 V by 250 ohms (125 ohms * 2), or is the total resistance 125 ohms so the current would be 50V divided by 125 ohms? This is the part I’m stuck on, seeing how each method is covered by the answer choices. Problem 4. (4 points) LC circuits. In the circuit of Fig. 3, the battery emfis 50 V, the resistance is 12552, and the capacitance is 4.5 F. The switch s is closed for a long time interval, and zero potential difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum value of 50 V. What is the value of the inductance? (A) 35 mH (B) 70 mlH (C) 140.5 mH (D) 281 mH Figure 3. LC circuit. Problem 5. (4 points) Oscillations in LC circuits. What is the frequency of oscillation of the circuit in Problem 4 (A) 142 Hz (B) 284 Hz (C) 445 Hz (D) 890 Hz

Explanation / Answer

maximum current just before opening the switch is given as

i = E/(2R) = 50/(2 x 125) = 0.2 A

V = potential difference across capacitor = 50 volts

C = capacitance = 4.5 x 10-6 F

using conservation of energy

(0.5) L i2 = (0.5) C V2

L (0.2)2 = (4.5 x 10-6) (50)2

L = 0.28125 = 281 mH

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