In the oscillatory circuit, consisting of a capacitor with a capacity of C = 2 μF and a coil with an inductance

| June 21, 2021 | education

In the oscillatory circuit, consisting of a capacitor with a capacity of C = 2 μF and a coil with an inductance of L = 3 mH, harmonic oscillations occur. The amplitude value of the capacitor charge qmax = 6 μC. At a certain point in time, the current in the circuit is I = 24mA. Find the charge of the capacitor q at this moment.

To calculate the value of the charge of the capacitor used at the moment in time, we use the equality: Wm = WL + WC and qm ^ 2 / 2C = L * I ^ 2/2 + q ^ 2 / 2C, whence we express: q = √ (qm ^ 2 / C – L * I2) * C = √ (qm ^ 2 / C – L * I ^ 2) * C) = √ (qm ^ 2 – L * I2 * C).

Variables: qm – the amplitude value of the charge (qm = 6 μC = 6 * 10 ^ -6 C); L is the inductance of the coil (L = 3 mH = 3 * 10 ^ -3 H); I is the current at the moment in question (I = 24 mA = 24 * 10 ^ -3 A); C is the capacitance of the capacitor (C = 2 μF = 2 * 10 ^ -6 F).

Calculation: q = √ (qm2 – L * I ^ 2 * C) = √ ((6 * 10-6) ^ 2 – (3 * 10 ^ -3) * (24 * 10 ^ -3) ^ 2 * ( 2 * 10 ^ -6)) = 5.7 * 10 ^ -6 Cl.

Answer: The charge of the capacitor at the moment in question is 5.7 μC.



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