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An electric lamp which runs at 100V DC and consumes 10A current is connected to AC mains at 150V, 50Hz cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of the choke.

Answer
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Hint: Use the relation between the voltage drop across resistor and the voltage drop across inductor. Use the relation between the reactance, frequency and current to determine the inductance of the choke.

Formula used:
The voltage drop V is given by
V2=VL2+VR2 …… (1)
Here, VR is the voltage drop across the resistor and VL is the voltage drop across the inductor.
The voltage drop VL across inductor is given by
VL=IXL …… (2)
Here, I is the current and XL is the reactance.
The inductive reactance is given by
XL=2πfL …… (3)
Here, f is the angular frequency and L is the inductance.

Complete step by step answer:
An electric lamp which runs at 100V DC and consumes 10A current is connected to AC mains at 150V, 50Hz cycles with a choke coil in series.
Calculate the drop of voltage VL across the choke.
Rearrange equation (1) for the drop across inductor.
VL=V2VR2
Substitute 150V for V and 100V for VR in the above equation.
VL=(150V)2(100V)2
VL=2250010000
VL=12500
VL=111.8V
Hence, the voltage drop across the choke is 111.8V.
Determine the inductance of the choke.
Substitute 2πfL for XL in equation (2).
VL=I2πfL
VL=2πIfL
Rearrange the above equation for the inductance L.
L=VL2πIf
Substitute 111.8V for VL, 3.14 for π, 10A for I and 50Hz for f in the above equation.
L=111.8V2(3.14)(10A)(50Hz)
L=0.0356H
L=0.036H
Hence, the inductance is 0.036H.
Hence, the inductance and drop of voltage across the choke are 0.036H and 111.8V respectively.

Note:
If the resistance of the choke is not neglected, one may get the results other than the obtained.
Also note that you have to remember all the formulas and how to apply the formula in the specific numericals.