A coil has resistance 30 ohm and inductive reactance 20 Ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be :
Question
A coil has resistance 30 ohm and inductive reactance 20 Ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be :
Solution
To find the current in the coil, we first need to find the impedance (Z) of the coil. The impedance of a coil is given by the formula:
Z = sqrt(R^2 + X_L^2)
where R is the resistance and X_L is the inductive reactance.
Given that R = 30 ohm and X_L = 20 ohm, we can substitute these values into the formula:
Z = sqrt((30)^2 + (20)^2) Z = sqrt(900 + 400) Z = sqrt(1300) Z = 36.06 ohm
However, the frequency has been doubled from 50 Hz to 100 Hz. This will double the inductive reactance (since inductive reactance X_L = 2πfL, where f is the frequency and L is the inductance). So, the new inductive reactance is 2 * 20 = 40 ohm.
Now, we can calculate the new impedance with the new inductive reactance:
Z = sqrt((30)^2 + (40)^2) Z = sqrt(900 + 1600) Z = sqrt(2500) Z = 50 ohm
Now, we can find the current (I) using Ohm's law, which states that I = V / Z, where V is the voltage. Given that V = 200 volt, we can substitute these values into the formula:
I = 200 / 50 I = 4 Ampere
So, the current in the coil will be 4 Ampere.
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