(Inductance (L) = 0.9H, Capacitance (C) = 2×10−6
0.0009Ω/π
100Ω
1030Ω
400Ω
2500Ω
Correct answer is D
Given Data: Frequency (F) = \frac{500}{π} , Inductance (L) = 0.9H, Capacitance (C) = 2 \times 10^{-6}
Total circuit reactance = Inductive reactance ( X_L ) - Capacitive reactance ( X_C )
when ( X_L ) > ( X_C )
Inductive reactance ( X_L ) = 2πFL = 2 \times π \times \frac{500}{π} \times 0.9 = 900Ω
Capacitive reactance ( X_C ) = \frac{1}{2πFC} = \frac{1}{2 \times π \times 500/π \times 2 \times 10^{-6}}
= \frac{1}{2 \times 10^{-3}} = \frac{1}{0.002}
= 500Ω
Total circuit reactance = ( X_L ) - ( X_C ) = (900 - 500)Ω
=400Ω
8.00 x 106NC-1
4.00 x 106NC-1
4.00 x 10-4NC-1
2.00 x 10-4NC-1
2.50 x 10-1NC-1
Correct answer is B
F = qE; E = F/q = 40/(1.0 x 10-5) = 4.0 x 106
What is the value of R when G shows no deflection in the circuit illustrated above?
80Ω
75Ω
45Ω
20Ω
10Ω
Correct answer is D
R/30 = 100/150R = 20Ω
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