The net impedance of circuit (as shown in figure) will be [2023]

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Solution

Q.
The net impedance of circuit (as shown in figure) will be [2023]

1 $5 \sqrt{5}$ $Ω$

2 $25$ $Ω$

3 $10 \sqrt{2}$ $Ω$

4 $15$ $Ω$

Ans.
(1)

Given, $L = \frac{50}{π}$ $mH = \frac{50}{π} \times 10^{- 3}$ $H$

$f = 50$ Hz

Then $X_{L} = 2 π f L = 2 π \times 50 \times \frac{50}{π} \times 10^{- 3} = 5 Ω$ ...(i)

Similarly, $X_{C} = \frac{1}{2 π f C}$

Here, $C = \frac{10^{3}}{π}$ $μ F = \frac{10^{3}}{π} \times 10^{- 6}$ $F$

Then, $X_{C} = \frac{1}{2 π \times 50 \times \frac{10^{3}}{π} \times 10^{- 6}} = 10$ $Ω$

and $R = 10 Ω$

Therefore, net impedance $Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}} = \sqrt{10^{2} + {(10 - 5)}^{2}} = 5 \sqrt{5}$ $Ω$

Hence, option (1) is correct.

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