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  • A 150  resistor and an inductance L are connected in series to a 200 V, 50 Hz source of negligible impedance.

A 150 \Omega resistor and an inductance L are connected in series to a 200 V, 50 Hz source of negligible impedance. The current comes to 1 A. When the source is changed to 400 V, 100 Hz, the current will be:

Option: 1

less than 1 A


Option: 2

1 A


Option: 3

between 1 A and 2 A


Option: 4

between 2 A and 4 A


Answers (1)

best_answer

As \mathrm{I=\frac{200}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}}=1, \Rightarrow \sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}=200 }or

\mathrm{ \quad \mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2=40,000 }   or  \mathrm{(150)^2+X_L^2=40,000 }
or            \mathrm{\mathrm{X}_{\mathrm{L}}^2=40,000-22,500=17,500 Since v^{\prime}=100 \mathrm{~Hz} and v=50 \mathrm{~Hz}, \Rightarrow v^{\prime}=2 v }
\mathrm{As \quad X_L=2 \pi v L \Rightarrow \frac{X_L^{\prime}}{X_L}=\frac{2 \pi v^{\prime} L}{2 \pi v L}=2 \quad or \quad X_L^{\prime}=2 X_L or \quad \mathrm{X}_{\mathrm{L}}^{\prime}=4 \mathrm{X}_{\mathrm{L}}^2=4(17500)=70,000 }
\mathrm{\begin{aligned} & \therefore \mathrm{Z}^{\prime}=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^{\prime}}=\sqrt{(150)^2+70,000} \Omega=304 \Omega \\ & \therefore \mathrm{I}^{\prime}=\frac{\mathrm{V}^{\prime}}{\mathrm{Z}^{\prime}}=\frac{400}{304}=1.3 \mathrm{~A} \quad\left[\text { Given } \mathrm{V}^{\prime}=400 \mathrm{~V}\right] \end{aligned} }
\mathrm{(between 1 \mathrm{~A} \, and \, 2 \mathrm{~A} )}

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