Question

In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3\(\Omega\), 8\(\Omega\) and 4\(\Omega\) respectively, calculate phase difference between voltage and current.

Hint:

Remember {ELI the ICE man}. In an {L} (inductive) circuit, Voltage ({E}) leads Current ({I}). In a {C} (capacitive) circuit, Current ({I}) leads Voltage ({E}). Since \(X_L>X_C\) here, the circuit acts inductive (ELI).

Answer 1

The phase difference (\(\phi\)) between the voltage and current in a series LCR circuit is given by the formula: \[ \tan \phi = \frac{X_L - X_C}{R} \] Given:
Resistance, \(R = 3 \, \Omega\)
Inductive reactance, \(X_L = 8 \, \Omega\)
Capacitive reactance, \(X_C = 4 \, \Omega\)
Substituting the values: \[ \tan \phi = \frac{8 - 4}{3} = \frac{4}{3} \] Therefore, the phase difference is: \[ \phi = \tan^{-1}\left(\frac{4}{3}\right) \] Since \(X_L>X_C\), the circuit is inductive, and the voltage leads the current by an angle of \(\tan^{-1}(4/3)\), which is approximately \(53.13^{\circ}\).