Physics Hw Problem 3

# Physics Hw Problem 3 - Ω In Series 8 Ω 12 Ω = 20 Ω...

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Andy Gawne PHYS 133-51 Problem: The capacitors are charged and the switch closes at t = 0s. At what time has the current in the 8 resistor decayed to half the value it had immediately after the switch was closed? Model: We assume that the wire is ideal; there is no resistance that needs to be accounted for due to the wire. We also assume that the capacitors discharge through the resistors once the switch is closed. Visualize: Using the rules of capacitors and resistors in series and parallel, we can reduce the original circuit to a one-capacitor and one-resistor circuit. Capacitors: In Series: 1/60 + 1/60 = 1/C 1 ; C 1 = 30 μ f In Parallel: C 1 + 20 μ f = 30 μ f + 20 μ f = 50 μ f Resistors: In Parallel: 1/30 + 1/20 = R 1 ; R 1 = 12

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Unformatted text preview: Ω In Series: 8 Ω + 12 Ω = 20 Ω Andy Gawne PHYS 133-51 Solve: The time constant ( τ ) of the RC circuit is: τ = RC = (20 Ω ) * (50 μ f) = 1.0 ms Using the equation for RC circuits V = V o e-t/ τ and solving for t we get t = - τ * ln(V / V o ) Since Voltage = Current * Resistance and the resistance is constant while the current is halved, then V = V o /2. Substituting V = V o /2 into the equation t = - τ * ln(V / V o ) gives us: t = - τ * ln(1/2) = -1.0 ms * ln(1/2) = 0.69 ms Assess: The current through the 8 Ω resistor is half the initial current in 0.69 ms. This is a reasonable answer since the capacitance is such a small number and the resistor is fairly large in comparison....
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Physics Hw Problem 3 - Ω In Series 8 Ω 12 Ω = 20 Ω...

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