Question

1 What is the resistance of the inductive coil
takes 5A current across 240V, 50Hz supply
at 0.8 power factor?
A 48
B 4250
C 38.40
D 26.60​

Answer 1

Answer:

42.50

Explanation:

Answer 2

Inductive Resistance of the coil is (B) 42.5 Ω

Given:

Current in the circuit $(I)$  $=5A$  

Potential difference $(V)$ $=240V$

Frequency $(f)$ $=50Hz$

Power factor  $=0.8$

To find:

Inductive reactance $(X_{L} )$

Solution:

• We know, in a circuit, if only resistance is present, then the voltage and current in the circuit are always in the same phase.

                            $V_{R}= V_{o}cos$∅           $(i)$

       Where, $V_{o}$ is the peak voltage and $cos$∅ is the power factor

• And, in an inductive circuit, the respective voltage and current will be in a phase difference of $90^{o}$.

• Hence, in an LR circuit, the phase difference between the resistance and inductance will be $90^{o}$.

                           $V_{L}=V_{o}$ $cos$ ($90^{o}$ $+$  ∅)    $;$ $or$

                           $V_{L}=V_{o}sin$ ∅           $(ii)$

Step 2

We have,

$sin^{2} \alpha +cos^{2} \alpha =1$

Hence, $sin^{}\alpha = \sqrt{1-cos^{2}\alpha }$

We know, $cos$∅ $=0.8$   ;

Substituting this value in the equation, we get

$sin$∅$=0.6$

Step 3

Now, We know, $V_{rms} =\frac{V_{o} }{\sqrt{2} }$

We have the $rms$ value of voltage $(V_{rms} =240V)$, hence,

The peak value of voltage $V_{o}$ $={240}{\sqrt{2} } V$

Substituting the known values in $(ii)$, we get

$V_{L}= V_{o}sin$∅

$V_{L} = 240}{\sqrt{2} }(0.6)$

$V_{L} = 203.6V$

Hence, the potential difference through Inductor will be $203.6V$.

Step 4

Now,

According to Ohm's law, we have

$V=I}{X_{L} }$

Substituting the given values in the equation, we get

$X_{L}=\frac{V}{I}$

$X_{L}=\frac{203.6}{5}$

$X_{L}=42.5$ Ω $(approx)$

Final answer:

Hence, the resistance of the Inductive coil is $(B)42.5$.