Question

When a resistance R is connected in series with an element A, the electric current is found to be lagging behind the voltage by angle θ₁. When the same resistance is connected in series with element B, current leads voltage by θ₂. When R, A, B, are connected in series, the current now leads voltage by θ. Assume same AC source in used in all cases. Then:

(A) θ = θ₁ − θ₂

(B) tanθ = tanθ₂ − tanθ₁ 

(C) \(\theta = \frac{\theta _1 + \theta_2}{2}\)

(D) None of these

Answer 1

Correct option is (B) tanθ = tanθ₂ − tanθ₁

Let Z_A be the Impedance of element A, and Z_B be that of element B. Initially; when R is connected to A;

Given that current is lagging behind voltage by angle ‘θ₁ ’

When R is connected to B

Given that current leads voltage by ‘θ₂ ’

Using same method, when R, A, B are connected,