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In an RL series circuit, the current flowing is 5 A, the voltage drop across R is 16 V and the voltage drop across L is 12 V. Then what is the impedance?
  • a)
    2.4 Ω
  • b)
    4 Ω
  • c)
    5.6 Ω
  • d)
    3.2 Ω
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In an RL series circuit, the current flowing is 5 A, the voltage drop ...
Given that, I = 5A, VR = 16 V, VL = 12 V
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In an RL series circuit, the current flowing is 5 A, the voltage drop ...

Impedance Calculation in RL Series Circuit:


Given values:
- Current (I) = 5 A
- Voltage drop across resistor (R) = 16 V
- Voltage drop across inductor (L) = 12 V

Impedance Formula:
Impedance (Z) = √((R^2) + (XL^2))

Calculations:
- Resistance (R) = Voltage drop across resistor (V_R) / Current (I) = 16 V / 5 A = 3.2 Ω
- Inductive reactance (XL) = Voltage drop across inductor (V_L) / Current (I) = 12 V / 5 A = 2.4 Ω
- Impedance (Z) = √((3.2^2) + (2.4^2)) = √(10.24 + 5.76) = √16 = 4 Ω

Therefore, the impedance of the RL series circuit is 4 Ω. Hence, the correct answer is option 'B'.
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