In a parallel RL circuit, there are 3 A rms in the resistive branch and 3 A rms in the inductive branch. The total rms current is :

424 mA

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6 A

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4.24 A

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42.4 A

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Solution

The correct option is C 4.24 A
$I = \sqrt{I_{R}^{2} + I_{L}^{2}}$

where $I_{R}$ =current through resistor branch
and $I_{L}$ =current through inductive branch

Hence, $I = \sqrt{3^{2} + 3^{2}}$

$⟹ I = 3 \sqrt{2} A$

$⟹ I_{r m s} = 4.24 A$

Answer-(C)

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In a parallel RL circuit, there are 3 A rms in the resistive branch and 3 A rms in the inductive branch. The total rms current is :

The correct option is C 4.24 AI=√I2R+I2Lwhere IR=current through resistor branch and IL=current through inductive branchHence,I=√32+32⟹I=3√2A⟹Irms=4.24AAnswer-(C)

The correct option is C 4.24 A
$I = \sqrt{I_{R}^{2} + I_{L}^{2}}$

where $I_{R}$ =current through resistor branch
and $I_{L}$ =current through inductive branch

Hence, $I = \sqrt{3^{2} + 3^{2}}$

$⟹ I = 3 \sqrt{2} A$

$⟹ I_{r m s} = 4.24 A$

Answer-(C)