Two impedances $Z_{1} = (20 + j 10) Ω$ and $Z_{2} = (10 - j 30) Ω$ are connected in parallel and this combination is connected in series with $Z_{3} = 30 + j X .$ The value of $X$ which will produce resonance is

3.86

Ω

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0.26

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Solution

The correct option is A 3.86 $Ω$
Total impedance is

$Z = Z_{3} + (Z_{1} | | Z_{2})$

$= (30 + j X) + (\frac{(20 + j 10) (10 - j 30)}{20 + j 10 + 10 - j 30})$

$= 30 + j X + \frac{500 - j 500}{30 - j 20}$

$= 30 + j X + 19.23 - j 3.846$

At resonance, the imaginary part is zero.
$∴ X = 3.856 Ω \approx 3.86$ $Ω$

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Two impedances Z1=(20+j10)Ω and Z2=(10−j30)Ω are connected in parallel and this combination is connected in series with Z3=30+jX. The value of X which will produce resonance is

The correct option is A 3.86 ΩTotal impedance is Z=Z3+(Z1 || Z2) =(30+jX)+((20+j10)(10−j30)20+j10+10−j30) =30+jX+500−j50030−j20 =30+jX+19.23−j3.846 At resonance, the imaginary part is zero. ∴ X=3.856 Ω≈3.86 Ω

Two impedances $Z_{1} = (20 + j 10) Ω$ and $Z_{2} = (10 - j 30) Ω$ are connected in parallel and this combination is connected in series with $Z_{3} = 30 + j X .$ The value of $X$ which will produce resonance is