1. To find the effective value of the current, we need to first calculate the impedance of the coil.
The impedance of the coil Z can be found using the formula:
Z = R + jωL - 1/(jωC), where R is the resistance, L is the inductance, C is the capacitance, and ω = 2πf is the angular frequency. Since this is an AC circuit, we can assume that C is negligible (i.e., a series circuit with a capacitor and inductor of equal magnitude and frequency will cancel each other out).
So, Z = R + jωL = 5 + j(157)(0.05) = 5 + j7.85 ohms.
The effective value of the current, Ieff, is given by:
Ieff = √(Emax^2 / 2Z), where Emax is the maximum voltage of the emf.
Emax = √(200^2 + 50^2) = 206.2 V (using phasor addition).
Substituting in the values for Z and Emax, we get:
Ieff = √(206.2^2 / 2(5 + j7.85)) = 23.5 A (rounded to one decimal place).
Therefore, the effective value of the current is 23.5 Amperes.
2. We can use the formula for an RL circuit to find the time elapsed before the current reaches the operating value:
I(t) = I0(1 - e^(-Rt/L)),
where I0 is the initial current, R is the resistance, L is the inductance, and t is the time elapsed.
Given:
I0 = 15 mA = 0.015 A,
R = 800 ohms,
L = 15 H,
V = 20 V.
We can use Ohm's law to find the initial voltage across the buzzer:
V = IR = (0.015 A)(800 ohms) = 12 V.
The voltage drop across the inductor at t = 0 is zero, so the initial voltage drop across the buzzer is 20 V - 12 V = 8 V.
Setting I(t) equal to the operating current of 15 mA, we get:
0.015 = 0.015(1 - e^(-800t/15)).
Simplifying, we get:
1 = 1 - e^(-800t/15),
e^(-800t/15) = 0,
-800t/15 = ln(0),
t = infinity.
This tells us that the current will never reach the operating value of 15 mA, since the time constant (L/R) is much larger than the time we are interested in. In other words, the buzzer acts like a high-pass filter, allowing only high-frequency signals to pass through.
3. The circuit will oscillate when the reactance of the inductor is equal to the reactance of the capacitor:
XL = ωL = 2πfL = 2π(50)(1.0) = 314 ohms,
XC = 1/(ωC) = 1/(2πfC) = 1/(2π(50)(100x10^-6)) = 3183 ohms.
Thus, the minimum resistance required to prevent oscillation is R = XC - XL = 2869 ohms (rounded to integer value).
4. The energy supplied to the coil is given by the formula:
W = (1/2)L(I^2), where L is the inductance and I is the current.
Given:
R = 10 ohms,
L = 15 H,
V = 100 V,
t = 5 s.
Using Ohm's law, we can find the current in the circuit:
I = V/R = 100/10 = 10 A.
Substituting in the values for L and I, we get:
W = (1/2)(15)(10^2) = 750 J.
Therefore, the energy supplied to the coil in the first 5 seconds is 750 Joules.
PH
