Questions: 83. Three pure inductance are connected are shown in Fig the equivalent reactance to replace this circuit is (a) 0.4 H (c) 1.2 H (b) 0.8 H (d) 1.6 H

Solution Steps

• The circuit has three inductors: 1.2 H, 0.6 H, and 0.4 H.
• The 1.2 H and 0.6 H inductors are in parallel.
• The resulting parallel combination is in series with the 0.4 H inductor.

• Use the formula for parallel inductance: \[ \frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} \] where \(L_1 = 1.2 \text{ H}\) and \(L_2 = 0.6 \text{ H}\). \[ \frac{1}{L_{\text{eq}}} = \frac{1}{1.2} + \frac{1}{0.6} = \frac{1}{1.2} + \frac{2}{1.2} = \frac{3}{1.2} = 2.5 \] \[ L_{\text{eq}} = \frac{1}{2.5} = 0.4 \text{ H} \]

• The equivalent inductance of the parallel combination (0.4 H) is in series with the 0.4 H inductor.
• For series inductors, simply add the inductances: \[ L_{\text{total}} = L_{\text{eq}} + L_3 = 0.4 \text{ H} + 0.4 \text{ H} = 0.8 \text{ H} \]

Final Answer