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Mathematics
  1. Two 10 H 10 H inductors are in parallel and the parallel combination is in series with a third 10 H 10 H inductor. What is the approximate total reactance when a voltage with a frequency of 7 k H z 7kHz is applied across the circuit terminals?
    A. 219 k Ω 219 k Omega
    B. 66 k Ω 66 k Omega∼
    C. 660 k Ω 660 k Omega
    D. 1.3 M Ω quad1.3 M Omega
  2. Five inductors are connected in series. The lowest value is 8 μ H 8mu H. If the value of each inductor is twice that of the preceding one, and if the inductors are connected in order of ascending values, the total inductance is
    A. 8 H quad8H
    B. 32 μ H quad32 mu H
    C. 64 μ H quad64 mu H
    D. 248 μ H quad248 mu H
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  1. The equivalent inductance of the parallel combination of the two 10 H inductors is:
L eq = L 1 L 2 L 1 + L 2 = ( 10 H ) ( 10 H ) 10 H + 10 H = 5 H L_("eq")=(L_(1)L_(2))/(L_(1)+L_(2))=((10" H")(10" H"))/(10" H"+10" H")=5" H"
The total inductance of the circuit is:
L total = L 1 + L 2 + L 3 = ( 10 + 10 + 10 ) H = 30 H L_("total")=L_(1)+L_(2)+L_(3)=(10+10+10)"H"=30"H"
The impedance of an inductor is given by:
X L = ω L X_(L)=omega L
At the given frequency of 7 kHz, the reactance of each inductor is:
X L = ( 2 π × 7000 rad/s ) ( 10 H ) = 140 , 000 Ω = 140 k Ω X_(L)=(2pi xx7000" rad/s")(10" H")=140,000 Omega=140"k"Omega
The total reactance of the circuit is the sum of the three individual reactances:
X total = X L 1 + X L eq + X L 3 = 140 k Ω + 70 k Ω + 140 k Ω = 350 k Ω X_("total")=X_(L_(1))+X_(L_("eq"))+X_(L_(3))=140"k"Omega+70"k"Omega+140"k"Omega=350"k"Omega
Therefore, the approximate total reactance for the given circuit is 350 kΩ.
Answer: None of these (the options provided do not include 350 kΩ)
4. The values of the five inductors are:
L 1 = 8 μ H L_(1)=8mu"H"
L 2 = 2 L 1 = 16 μ H L_(2)=2L_(1)=16 mu"H"
L 3 = 2 L 2 = 32 μ H L_(3)=2L_(2)=32 mu"H"
L 4 = 2 L 3 = 64 μ H L_(4)=2L_(3)=64 mu"H"
L 5 = 2 L 4 = 128 μ H L_(5)=2L_(4)=128 mu"H"
The total inductance of the circuit is the sum of the individual inductances:
L total = L 1 + L 2 + L 3 + L 4 + L 5 L_("total")=L_(1)+L_(2)+L_(3)+L_(4)+L_(5)
Substituting the given values, we get:
L total = ( 8 μ H ) + ( 16 μ H ) + ( 32 μ H ) + ( 64 μ H ) + ( 128 μ H ) L_("total")=(8mu"H")+(16 mu"H")+(32 mu"H")+(64 mu"H")+(128 mu"H")
Simplifying, we get:
L total = 248 μ H L_("total")=248 mu"H"
Therefore, the total inductance of the circuit is approximately 248 µH.
Answer: D. 248 μ H 248 mu H
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