Question

Как измениться период свободных колебаний в контуре с ростом электроемкости конденсатора в 25 раз и уменьшении индуктивности в 4 раза?

Answer 1

T1=2*pi*sqrt(L*C)
T2=2*pi*sqrt(25L*C/4)
T2/T1=5/2=2,5
T2=2,5 T1 увеличится в 2,5 раза

Answer 2

Changing the Free Oscillation Period in an Electrical Circuit

To change the period of free oscillations in a circuit with a 25-fold increase in the capacitance of the capacitor and a 4-fold decrease in inductance, we can use the formula for the period of oscillation in an LC circuit, which is given by:

T = 2π(LC)^0.5

Where: - T = period of oscillation - L = inductance - C = capacitance

Calculation

1. Initial Period of Oscillation (T_initial): - Let's assume the initial values of inductance (L_initial) and capacitance (C_initial) are 1 unit each. - Using the formula, the initial period of oscillation (T_initial) can be calculated as follows: - T_initial = 2π(1*1)^0.5 - T_initial = 2π

2. New Period of Oscillation (T_new): - With a 25-fold increase in capacitance (C_new = 25) and a 4-fold decrease in inductance (L_new = 1/4), the new period of oscillation (T_new) can be calculated as follows: - T_new = 2π((1/4)*25)^0.5 - T_new = 2π(25/4)^0.5 - T_new = 2π(5/2) - T_new = 5π

Conclusion

With a 25-fold increase in the capacitance of the capacitor and a 4-fold decrease in inductance, the period of free oscillations in the circuit will increase to 5π from the initial period of 2π. This means the period will increase by approximately 2.5 times.

[[6 #]], [[7 #]]