An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ' R' are connected in series to an ac source of potential difference ' V volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2 A. The impedance of the circuit is:

Option 1 -

4√2 Ω

Option 2 -

5/√2 Ω

Option 3 -

4 Ω

Option 4 -

5 Ω

3 Views | Posted 4 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 weeks ago

Correct Option - 2

Detailed Solution:

Diameter = main scale reading + (circular scale reading × least count)
Diameter = 0 + (52 × 0.01 mm) = 0.52 mm = 0.052 cm.
In the RLC circuit:
Given I? = 10√2 A, so I? = I? /√2 = 10 A.
V? = √ [V? ² + (V? - V? )²] = √ [40² + (40 - 10)²] = √ [1600 + 30²] = √ [1600 + 900] = √2500 = 50 V.
Impedance Z = V? / I? = 50 V / 10 A = 5 Ω.
For Hindi: I? = 10√2 A, V? = 50V, Z = V? /I? = 50/ (10√2) = 5/√2 Ω.

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