The figure below shows a circuit with em f = 5.00 V, R = 6.10 Ω, L = 76.0 mH, and C = 2.60 µF. After a long time interval at the position a shown in the figure, the switch S is thrown to position b at time t = 0.
(a) What is the maximum charge on the capacitor? (Enter the magnitude.)
(b) What is the maximum current in the inductor for
t > 0? (Enter the magnitude.)
(c) What is the frequency of oscillation of the resulting LC circuit for
t > 0?
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To find the maximum charge on the capacitor in the given circuit, we need to use the formula for the charge on a capacitor:
Q = CV
where Q is the charge, C is the capacitance, and V is the voltage.
In this case, the circuit has a capacitance of C = 2.60 µF and a voltage of V = 5.00 V. Plugging these values into the formula:
Q = (2.60 µF) * (5.00 V) = 13.0 µC
So, the maximum charge on the capacitor is 13.0 µC.
To find the maximum current in the inductor for t > 0, we can use the formula for the current in an inductor:
I = V / L * t
where I is the current, V is the voltage, L is the inductance, and t is time.
In this case, the circuit has an inductance of L = 76.0 mH and a voltage of V = 5.00 V. Since we are interested in the maximum current, we can assume that t is sufficiently large. Plugging these values into the formula:
I = (5.00 V) / (76.0 mH) * t
The maximum current will occur at t = infinity (after a long time interval), so as t approaches infinity, the current will approach zero.
To find the frequency of oscillation of the resulting LC circuit for t > 0, we can use the formula:
f = 1 / (2π√(LC))
where f is the frequency, L is the inductance, and C is the capacitance.
In this case, the circuit has an inductance of L = 76.0 mH and a capacitance of C = 2.60 µF. Plugging these values into the formula:
f = 1 / (2π√(76.0 mH * 2.60 µF))
Simplifying using the appropriate units:
f = 1 / (2π √(0.076 H * 2.60 * 10^(-6) F))
f ≈ 61.2 Hz
So, the frequency of oscillation of the resulting LC circuit for t > 0 is approximately 61.2 Hz.