Question

5. A non-inductive resistance of 10 Ω is connected in series with an inductive coil of 2 Ω resistance. The combination is then supplied by 200 V, 50 Hz AC source. If the current drawn by the circuit is 10 A then calculate the followings:

(a) Inductance of coil (b) Power Factor (c) Voltage across Coil

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Answer 1

Answer:

(a)Inductance-$\frac{1}{50\pi}$

(b)Power factor- $\frac{10}{\sqrt{104} }$

(c)Voltage across the coil-  $100V$

Explanation:

The values given in the question are,

R=10Ω

XL=2Ω

V=200v

Frequency(f)=50Hz

I=10A

(a)

Inductive reactance is given as,

$X_L=\omega L$            (1)

XL=inductive reactance

ω=angular frequency=2πf

L=inductance

Now by substituting the value of XL and ω in equation (1) we get;

$2=2\pi\times 50\times L$

$L=\frac{1}{50\pi}$              (2)

(b)

The power factor of the circuit is given as,

$cos\phi=\frac{R}{Z}$      (3)

Where,

cosФ=power factor

R=resistance of the circuit

Z=impedence of the circuit

Impedance is given as,

$Z=\sqrt{R^2+X^2_L}$         (4)

By substituting the value of R and XL in equation (4) we get;

$Z=\sqrt{10^2+2^2}=\sqrt{100+4}=\sqrt{104}$       (5)

So, the power factor of the circuit is given as,

$cos\phi=\frac{10}{\sqrt{104} }$

(c)

As the resistance and inductive coil are connected in series the current through them will be the same. So, first, let's find the voltage across the resistance i.e.

$V_R=10\times 10=100V$     (6)

Now the voltage across the inductive coil is given as,

$V_L=V-V_R$                   (7)

By substituting the value of V and VR in equation (7) we get;

$V_L=200-100=100V$

Hence, the inductance of the coil is $\frac{1}{50\pi}$, the power factor is $\frac{10}{\sqrt{104} }$ and the voltage across the coil is $100V$.