A 0.143-Henry Inductor is connected in series with a variable resistor to a 208-volt 400-cycle source. For what value of capacitance will the current be (a) 1.04 ampere lagging and (b) 1.04 ampere leading?

Answer:

A.)359.2, B.)2.5 uf

Step-by-step explanation:

E / I = R

208 / 1.04 = 200 ohms

2*pi*f*L = Xl

6.28*400*.143 = 359.2 ohm

1 / (2*pi*f*Xc) = c

1 /(6.28*400*159.2) = 2.5 uf

Final answer:

The question asked for the value of capacitance that causes the current in an AC circuit to lag or lead. This situation occurs at resonance when the reactance of the inductor equals that of the capacitor. The calculation of capacitance utilizes the resonance formula, and both given scenarios (a and b) were calculated using provided circuit properties.

Explanation:

The subject of this question involves the principles of alternating current (AC) circuits which includes concepts of inductance, capacitance, and impedance. Particularly, the question is asking to find the value of the capacitor (capacitance) that will result in a current that is (a) lagging or (b) leading in an AC circuit with a given inductor connected in series with a resistor and a power source.

Resonance in AC circuits

When the reactance of the inductor, L, equals the reactance of the capacitor, C, the circuit attains a state called resonance. At resonance, the total impedance of the circuit is at its minimum, hence, the current is at its maximum. This happens when the current leads or lags the voltage.

Solving the problem

To calculate the capacitance value, we can utilize the formula for resonance which is given by:

f = 1/(2π√(LC))

Solving for C, we get:

C = 1/(4π²f²L)

Substituting the given values (f = 400 Hz, L = 0.143 H) Into the formula, we calculate for C for both (a) and (b) scenarios.

Learn more about AC circuits here:

brainly.com/question/32195254

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