Question

When a resistance R is connected in series with a non-resistive element A, the electric current is found to be lagging behind the voltage V by angle 30°. When the same resistance is connected in series with a non-resistive element B, current leads by 60°. When R, A, B are connected in series, the current now leads voltage by $\theta$ which is equal to tan$^{-1}$ $(K/\sqrt{3})$, then the value of K is (assume same AC source is used in all cases).

Answer 1

2

Answer 2

The problem involves analyzing the phase relationship between current and voltage in different AC series circuits. We will use the concept of impedance and phase angle for RL, RC, and RLC circuits.

Step 1: Analyze the circuit with Resistance R and non-resistive element A.

• The current lags behind the voltage by $30^\circ$. This indicates that element A is an inductor (L).
• Let the inductive reactance of element A be $X_A$.
• For an RL circuit, the phase angle $\phi_1$ is given by: $\tan \phi_1 = \frac{X_A}{R}$
• Given $\phi_1 = 30^\circ$: $\tan 30^\circ = \frac{X_A}{R}$ $\frac{1}{\sqrt{3}} = \frac{X_A}{R}$ $X_A = \frac{R}{\sqrt{3}}$

Step 2: Analyze the circuit with Resistance R and non-resistive element B.

• The current leads the voltage by $60^\circ$. This indicates that element B is a capacitor (C).
• Let the capacitive reactance of element B be $X_B$.
• For an RC circuit, the phase angle $\phi_2$ is given by: $\tan \phi_2 = \frac{X_B}{R}$
• Given $\phi_2 = 60^\circ$: $\tan 60^\circ = \frac{X_B}{R}$ $\sqrt{3} = \frac{X_B}{R}$ $X_B = R\sqrt{3}$

Step 3: Analyze the circuit with R, A, and B connected in series.

• This is an RLC series circuit, where A is an inductor ($X_A$) and B is a capacitor ($X_B$).
• The net reactance ($X_{net}$) of the circuit is the difference between inductive and capacitive reactances: $X_{net} = X_A - X_B$ $X_{net} = \frac{R}{\sqrt{3}} - R\sqrt{3}$ $X_{net} = R \left( \frac{1}{\sqrt{3}} - \sqrt{3} \right)$ $X_{net} = R \left( \frac{1 - 3}{\sqrt{3}} \right)$ $X_{net} = -\frac{2R}{\sqrt{3}}$
• The negative sign indicates that the net reactance is capacitive ($X_B > X_A$), which means the current leads the voltage. This is consistent with the problem statement "current now leads voltage by $\theta$".
• The phase angle $\theta$ for the RLC circuit is given by: $\tan \theta = \frac{|X_{net}|}{R}$ $\tan \theta = \frac{\left|-\frac{2R}{\sqrt{3}}\right|}{R}$ $\tan \theta = \frac{\frac{2R}{\sqrt{3}}}{R}$ $\tan \theta = \frac{2}{\sqrt{3}}$

Step 4: Determine the value of K.

• The problem states that $\theta = \tan^{-1} (K/\sqrt{3})$.
• Therefore, $\tan \theta = \frac{K}{\sqrt{3}}$.
• Comparing this with our calculated value of $\tan \theta$: $\frac{K}{\sqrt{3}} = \frac{2}{\sqrt{3}}$ $K = 2$

The final answer is $\boxed{2}$.

Explanation of the solution:

1. Identify element A as an inductor from the lagging current, and calculate its reactance $X_A = R/\sqrt{3}$ using $\tan 30^\circ = X_A/R$.
2. Identify element B as a capacitor from the leading current, and calculate its reactance $X_B = R\sqrt{3}$ using $\tan 60^\circ = X_B/R$.
3. For the series RLC circuit, calculate the net reactance $X_{net} = X_A - X_B = R/\sqrt{3} - R\sqrt{3} = -2R/\sqrt{3}$.
4. Since $X_{net}$ is negative, the circuit is capacitive, and current leads voltage. The phase angle $\theta$ is given by $\tan \theta = |X_{net}|/R = |-2R/\sqrt{3}|/R = 2/\sqrt{3}$.
5. Equate this to the given form $\tan \theta = K/\sqrt{3}$ to find $K=2$.