Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be:

$P(\frac{R}{Z})^2$

$P\sqrt{\frac{R}{Z}}$

$P(\frac{R}{Z})$

$P$

Step-by-Step Solution

Initial Case (Pure Resistor): When the AC source of voltage $V$ is connected to a pure resistance $R$ , the power drawn is given by $P = \frac{V^2}{R}$ . From this, we can express the source voltage squared as $V^2 = P R$ .
Final Case (Series RL Circuit): An inductance is placed in series with the resistance. The total impedance of the circuit is given as $Z$ . The new RMS current in the circuit is $I' = \frac{V}{Z}$ .
Power Calculation: In an AC circuit, average power is dissipated only by the resistive component (pure inductance consumes zero average power). The new power drawn is $P' = I'^2 R$ .
Substitution: Substituting the expression for the new current: $P' = (\frac{V}{Z})^2 R = \frac{V^2 R}{Z^2}$ .
Relation to P: Substitute $V^2 = P R$ from the initial case into the new power equation: $P' = \frac{(P R) R}{Z^2} = P \frac{R^2}{Z^2} = P(\frac{R}{Z})^2$ .

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