Question

In LCR alternating circuit , R=10 ohm ,XL= 100 ohm and Xc =100 ohm .write the value of impedance of circuit​

Answer 1

Given :

• Resistance (R) = 10 ohm
• Inductive Reactance $(\sf X_L)$ = 100 ohm
• Capacitive Reactance $(\sf X_C)$ = 100 ohm

To Find :

• Impedence of the circuit (Z)

Formula Applied :

In a series of LCR circuit, the impedence of circuit is given by,

$\underline{\boxed{\sf Z =\sqrt{ R^2+(X_L-X_c)^2}}}$

In where ,

• Z = Impedence
• R = Resistance
• $(\sf X_L)$ = Inductive Reactance
• $(\sf X_C)$ = Capacitive Reactance

Solution :

$\longrightarrow \sf Z = \sqrt{R^2+(X_L-X_C)^2}\\\\\longrightarrow \sf Z = \sqrt{10^2+(100-100)^2}\\\\\longrightarrow \sf Z = \sqrt{100+(0)^2}\\\\\longrightarrow \sf Z = \sqrt{100}\\\\\longrightarrow \sf Z = 10 Ω$

Required Answer :

The value of impedance of circuit is $\underline{\sf 10 Ω}$

Answer 2

Answer:

The value of the impedance of the circuit​ is 10 ohms.

Explanation:

Given that,

Resistance, R = 10 ohms

Inductive reactance, $X_{L}$ = 100 ohm

Capacitive reactance, $X_{C}$ = 100 ohm

To Calculate :
The value of the impedance of the circuit.

Calculations:

We need to find the impedance of the circuit. In a series LCR circuit, the impedance of the  circuit is given by :

$Z = \sqrt{R^{2} +(X_{L} -X_{C} )^{2} }$

Substituting all the values, we have

$Z = \sqrt{(10)^{2} +(100 -100 )^{2} }= 10$ ohm

This is a purely resistive circuit. With a circuit that is entirely resistive, all of the voltage used is used to overcome the ohmic resistance of the circuit. This is also the condition of resonance.

Therefore,  the value of the impedance of the circuit​ is 10 ohms.

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