Question

An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is

• A. π/6
• B. π/4
• C. π/2
• D. zero

Hint:

The phase difference formula is used to determine the phase difference between the current and applied voltage.

Answer 1

The Correct Option is B

\(\tan \phi^{-1}=\frac{ X _{ L }}{ R }\)

= tan^-1(3/3)

= tan^-1(1)

= π/4 rad

\(\phi=45^{\circ}\)

Hence, option B) π/4 is the correct answer.

Answer 2

The phase difference formula is used to determine the phase difference between the current and applied voltage.

The phase difference between the applied voltage and the current in the circuit can be calculated by,

\(tanϕ ={X_L\over R}\)

• Where, \(ϕ\) denotes the phase difference between the applied voltage and the current
• \(X_L\) is the inductive reactance 
• R is the applied resistance 

Given - Resistance of the circuit = 3Ω

The inductive reactance of the inductor X_L = 3Ω

tanϕ = X_L/R — 1)

Substituting inductive reactance of the inductor and the resistance in equation (1) - 

tan ϕ = 3/3

Therefore, tanϕ = 1

ϕ = tan^-1

From trigonometry, the value of the tan^-1(1) = 45^o

ϕ = 45^o

The angle 45° is equal to π/4 \(\frac{π}{4}\), so

ϕ = π/4 

Thus, the above equation shows the phase difference between the applied voltage and the current in the circuit.

Hence, option (B): \(\dfrac{\pi}{4}\) is the correct answer.

An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is π/4.