In a sinusoidally driven series RLC circuit, the inductive resistance is XL = 100 Ω, the capacitive reactance is XC = 200 Ω, and the resistance is R = 50 Ω. The current and applied emf would be in phase if

The current and the applied emf can be in phase if either of the two changes are made.

1) The inductance of the inductor is doubled, with everything else remaining constant.

2) The capacitance of the capacitor is doubled, with everything else remaining constant.

Explanation:

The current and applied emf for this type of circuit would be in phase when there is no phase difference between the two quantities. That is, Φ = 0°.

The phase difference between current and applied emf is given as

Φ = tan⁻¹ [(XL - Xc)/R]

XL = Impedance due to the inductor

Xc = Impedance due to the capacitor

R = Resistance of the resistor.

For Φ to be 0°, tan⁻¹ [(XL - Xc)/R] = 0

But only tan⁻¹ 0 = 0 rad

So, for the phase difference to be 0,

[(XL - Xc)/R] = 0

Meaning

XL = Xc

But for this question,

XL = 100 Ω, Xc = 200 Ω

For them to be equal, we have to find a way to increase the impedance of the inductor or reduce the impedance of the capacitor.

The impedance are given as

XL = 2πfL

Xc = (1/2πfC)

f = Frequency

L = Inductance of the inductor

C = capacitance of the capacitor

The impedance of the inductor can be increased from 100 Ω to 200 Ω by doubling the inductance of the inductor.

And the impedance of the capacitor can be reduced from 200 Ω to 100 Ω by also doubling the capacitance of the capacitor.

So, these are either of the two ways to make the current and applied emf to be in phase.

Hope this Helps!!!

the answer is a 3.8cm

explanation: