Two coils A and B are connected in series across a 240 V- 50 Hz supply- The resistance of A is 5 - and the inductance of B is 0-02 H- The power consumed is 3 kW and the power factor is 0-75- The impedence of the circuit is

The correct option is C 14.4 Ω Given : #160; E_v = 240 V , #160; cosϕ = 0.75 and p = 3000 W Power consumed #160; P=E_v^2cosϕ ...

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Standard XII

Physics

Galvanometer

Two coils A...

Question

Two coils $A$ and $B$ are connected in series across a $240 V, 50 Hz$ supply. The resistance of $A$ is $5 Ω$ and the inductance of $B$ is $0.02 H$ . The power consumed is $3 kW$ and the power factor is $0.75$ . The impedence of the circuit is

0.144 Ω

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1.44 Ω

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14.4 Ω

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144 Ω

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Solution

The correct option is C $14.4 Ω$
Given : $E_{v} = 240 V$ , $cos ϕ = 0.75$ and $p = 3000 W$
Power consumed $P = \frac{E_{v}^{2} cos ϕ}{Z} = 3000$
where $Z$ is the impedence of the circuit.
$∴ \frac{(240)^{2} (0.75)}{Z} = 3000$
$\Rightarrow Z = 14.4 Ω$

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Two coils A and B are connected in series across a 240 V,50 Hz supply. The resistance of A is 5 Ω and the inductance of B is 0.02 H. The power consumed is 3 kW and the power factor is 0.75. The impedence of the circuit is

The correct option is C 14.4 ΩGiven : Ev=240 V, cosϕ=0.75 and p=3000 WPower consumed P=E2vcosϕZ=3000where Z is the impedence of the circuit.∴(240)2(0.75)Z=3000 ⇒Z=14.4 Ω

Two coils $A$ and $B$ are connected in series across a $240 V, 50 Hz$ supply. The resistance of $A$ is $5 Ω$ and the inductance of $B$ is $0.02 H$ . The power consumed is $3 kW$ and the power factor is $0.75$ . The impedence of the circuit is

The correct option is C $14.4 Ω$
Given : $E_{v} = 240 V$ , $cos ϕ = 0.75$ and $p = 3000 W$
Power consumed $P = \frac{E_{v}^{2} cos ϕ}{Z} = 3000$
where $Z$ is the impedence of the circuit.
$∴ \frac{(240)^{2} (0.75)}{Z} = 3000$
$\Rightarrow Z = 14.4 Ω$