Answer
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Hint:Here, we are given two elements X and Y, where X is a pure resistive circuit element and Y is some element which is unknown. To find what kind of element Y is, consider the statement which describes the current voltage relation. From the given data and conditions, find the impedance and resistance of Y and X. Accordingly, combine the elements in series and then find the rms value of the current.
Complete step by step solution:
First, let us consider the element X. As it is given that it is pure resistive, the element X is a resistor.
The current will be given by the Ohm’s Law $ \to V = IR \to R = \dfrac{V}{I} = \dfrac{{200}}{5} =
40\Omega $
The resistance of the element X $40\Omega $.
Now, let us consider the second element Y. As it is given that the current lags the voltage by an angle
of $90^\circ $, the element must be pure inductive. Therefore, impedance that will be offered by Y will be ${X_L} = \dfrac{{200}}{5} = 40\Omega $
Combine the elements X and Y in series, we will get an LR circuit. The impedance will be given by $Z
= \sqrt {{R^2} + {X_L}^2} = \sqrt {{{40}^2} + {{40}^2}} = 40\sqrt 2 \Omega $
The peak current will be given by ${I_0} = \dfrac{{{V_0}}}{Z} = \dfrac{{200}}{{40\sqrt 2 }} =
\dfrac{5}{{\sqrt 2 }}A$.
The rms value of the current will be ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }} = \dfrac{{\dfrac{5}{{\sqrt
2 }}}}{{\sqrt 2 }} = \dfrac{5}{2}A$.
Hence, if the series combination of X and Y is connected to the same supply, the rms value of current
will be $\dfrac{5}{2}A$
Hence, the Correct Option is (C).
Note:Whenever an element is given, find what type of element is it from the given data or the statements about the element. When the circuit element is pure resistive, the element is a resistor, when the current lags the voltage by $90^\circ $, the element is an inductor and when the current leads the voltage by an angle of $90^\circ $, the element acts as a capacitor. Remember the formula for the impedance of the circuit which is the resultant of the impedances.
Complete step by step solution:
First, let us consider the element X. As it is given that it is pure resistive, the element X is a resistor.
The current will be given by the Ohm’s Law $ \to V = IR \to R = \dfrac{V}{I} = \dfrac{{200}}{5} =
40\Omega $
The resistance of the element X $40\Omega $.
Now, let us consider the second element Y. As it is given that the current lags the voltage by an angle
of $90^\circ $, the element must be pure inductive. Therefore, impedance that will be offered by Y will be ${X_L} = \dfrac{{200}}{5} = 40\Omega $
Combine the elements X and Y in series, we will get an LR circuit. The impedance will be given by $Z
= \sqrt {{R^2} + {X_L}^2} = \sqrt {{{40}^2} + {{40}^2}} = 40\sqrt 2 \Omega $
The peak current will be given by ${I_0} = \dfrac{{{V_0}}}{Z} = \dfrac{{200}}{{40\sqrt 2 }} =
\dfrac{5}{{\sqrt 2 }}A$.
The rms value of the current will be ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }} = \dfrac{{\dfrac{5}{{\sqrt
2 }}}}{{\sqrt 2 }} = \dfrac{5}{2}A$.
Hence, if the series combination of X and Y is connected to the same supply, the rms value of current
will be $\dfrac{5}{2}A$
Hence, the Correct Option is (C).
Note:Whenever an element is given, find what type of element is it from the given data or the statements about the element. When the circuit element is pure resistive, the element is a resistor, when the current lags the voltage by $90^\circ $, the element is an inductor and when the current leads the voltage by an angle of $90^\circ $, the element acts as a capacitor. Remember the formula for the impedance of the circuit which is the resultant of the impedances.
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