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Assoli18 [71]
3 years ago
12

A part of a circuit contains an inductor. The current through the inductor is changing uniformly from 2.40A to 0.30A over the co

urse of 1.75s. If the EMF ( voltage change ) from one side of the inductor to the other is 5.70 Volts, what is the value of the inductance
Physics
1 answer:
Nataly_w [17]3 years ago
6 0

Answer:

The value of the inductance is 4.75 H

Explanation:

Given;

initial current through the inductor, I₁ = 2.4 A

final current through the inductor, I₂ = 0.3 A

duration of change of current, dt = 1.75 s

voltage change of the inductor, V = 5.7 Volts

The voltage change of the inductor is given by;

V_L = -L\frac{di}{dt}\\\\ V_L = -L(\frac{I_2-I_1}{dt} )\\\\V_L = L(\frac{I_1-I_2}{dt} )\\\\

Where;

L is the inductance of the coil;

5.7 = L(\frac{2.4-0.3}{1.75} )\\\\5.7 = 1.2 L\\\\L = \frac{5.7}{1.2}\\\\ L = 4.75 \ H

Therefore, the value of the inductance is 4.75 H

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Read 2 more answers
An AC source of maximum voltage V0 = 30 V is connected to a resistor R = 50 Ω, an inductor L = 0.6 H, and a capacitor C = 20 µF.
ivolga24 [154]

Hello!

We can begin by solving for the resonance ANGULAR frequency of the circuit.

For an RCL circuit, the resonance angular frequency is given as:
\omega_0^2 = \frac{1}{LC}\\\\w_0 = \sqrt{\frac{1}{LC}}

ω₀ = resonance angular frequency (rad/s)

L = Inductance (0.6 H)
C = Capacitance (20 μF)

Plug in the values and solve.

\omega_0 = \sqrt{\frac{1}{(0.6)(0.00002)} } = 288.675 \frac{rad}{s}

For an AC power source, the output is usually expressed as:


V(t) = V_{max}sin(\omega_0 t})

So, using the appropriate values and setting the source angular frequency equivalent to the circuit's resonance angular frequency:


V(t) = 30sin(288.675t)

To find the maximum charge on the capacitor when the frequency of the source is equivalent to the resonance frequency of the circuit (or the angular frequencies are equal), we can begin by finding the maximum voltage across the capacitor.

To find this, however, we must solve for the maximum current across the circuit by finding the total impedance of the circuit. When the circuit is at resonance, the impedance is equivalent to the resistance of the RESISTOR.

So, solve for the maximum current in the circuit using Ohm's Law:

i = \frac{V}{R}

In this instance AT RESONANCE:

I_{Max} = \frac{V_Max}{R}\\\\I_{Max} = \frac{30}{50} = 0.6 A

Now, we must solve for the capacitive reactance in order to find the maximum voltage across the capacitor. Using the following equation for capacitive reactance:
X_c = \frac{1}{\omega C}\\\\X_c = \frac{1}{(288.675)(0.00002)} = 173.205 \Omega

Now that we found the maximum current and capacitive reactance, we can now solve for the maximum voltage across the capacitor:
V_{C, max} = X_C I_{Max}\\\\V_{C, max} = 173.205 * 0.6 = 103.923 V

Finally, we can easily solve for the maximum charge on the capacitor using the relationship:
C = \frac{Q}{V}\\\\Q = CV

Plug in the values solved for above.

Q = (0.00002)(103.923) = 0.00208 C = \boxed{2.078 mC}

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