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Question

A serial circuit with resistor of 600 ohm and capacity of 4μF is connected to the source of alternating circuit of 100 V with a voltage equation V=Vmsin 250t.The magnitude of inductor that must be serially installed to the circuit in order to flow the current of 100mA is...H

A.0.1
B.0.2
C.0.4
D.0.8
E.1

Answer 1

1. The current in the given circuit is 100mA which is exactly 0.1 A
2. In the equation of V= V_msin(250t) , here ω=250 rad/sec.
3. R = 600Ω and C = 4μF , and in Series LCR circuit we need to find L(Inductance).
4. Step 1 :- Find Impedance in the circuit, so z=V/i :- z = 1000 Ω.
5. Step 3 :- X_c= 1⁄ωC = (1⁄(250×4)))×10⁶= 10³Ω and X_L = ωL = 250L
6. Step 4 :- Use z = √(R²+(X_L-X_C)²), 1000 = √(600² + ( 250L-1000)²), 10⁶ = 600² + 62500L² + 10⁶- 5×10⁵×L, 0 = 360000 + 62500L² - 500000L, 0= 3600 + 625L² - 5000L
7. Step 5 :- Use x = (-b+or-√(b²-4×a×c))/(2×a) so, L = (5000+or-√(25×10⁶-9×10⁶))/1250, L = (5000+or-4000)/1250, L₁ = (5000+4000)/1250 = 7.2H, L₂ = (5000-4000)/1250 = 0.8H.
8. Step 6 :- Match with options and eliminate other unnecessary options out, now we have 7.2H and 0.8H. Because we do not have 7.2H and we have 0.8H in option d, Hence d is the correct answer. But it can be 7.2H scientifically. In fact, both 0.8H and 7.2H satisfy the circuit condition mathematically, because they give the same impedance magnitude and therefore the same current.