
A 220V, 50Hz AC generator is connected to an inductor and a $50\Omega $ resistance in series. The current in the circuit is 1.0A. What is the potential difference across the inductor?
A. 102.2V
B. 186.4V
C. 214V
D. 170V
Answer
529.8k+ views
Hint: You could make a circuit diagram for better clarity of the given situation. Then you could use Ohm’s law to find the net impedance of the circuit from the given value of source voltage and current. Then you could find the inductive reactance and then apply Ohm’s law again for the inductor alone and thus find the answer.
Formula used:
Ohm’s law,
$V=IZ$
Complete step by step solution:
In the question, we are given an AC generator that is connected in series with an inductor as well as resistance as shown in the figure. The resistance of the resistor is given as $50\Omega $ and the current in the circuit is found to be 1.0A.
By applying Ohm’s law in the above circuit we get,
$I=\dfrac{V}{Z}$
Where, Z is the impedance of the circuit.
$\Rightarrow Z=220\times 1=220\Omega $
We know that the impedance in the above circuit with inductor and resistor connected in series would be given by,
$Z=\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}$
$\Rightarrow {{X}_{L}}=\sqrt{{{Z}^{2}}-{{R}^{2}}}$
$\Rightarrow {{X}_{L}}=\sqrt{{{220}^{2}}-{{50}^{2}}}=214.24\Omega $
Since the connection is in series, the current will be the same 1A through the inductor and now we also have the inductive reactance. We could apply Ohm’s law here to get,
${{V}_{i}}=I{{X}_{L}}$
$\therefore {{V}_{i}}=1\times 214.24\approx 214V$
Therefore, we found the potential difference across the inductor to be 214V. Hence, option C is the correct answer.
Note: Impedance simply is the net opposition offered to the current in a particular circuit. The difference between impedance and resistance is that the impedance takes the effect of inductor and capacitor into account. The SI unit of impedance is also the same as resistance –ohms.
Formula used:
Ohm’s law,
$V=IZ$
Complete step by step solution:
In the question, we are given an AC generator that is connected in series with an inductor as well as resistance as shown in the figure. The resistance of the resistor is given as $50\Omega $ and the current in the circuit is found to be 1.0A.
By applying Ohm’s law in the above circuit we get,
$I=\dfrac{V}{Z}$
Where, Z is the impedance of the circuit.
$\Rightarrow Z=220\times 1=220\Omega $
We know that the impedance in the above circuit with inductor and resistor connected in series would be given by,
$Z=\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}$
$\Rightarrow {{X}_{L}}=\sqrt{{{Z}^{2}}-{{R}^{2}}}$
$\Rightarrow {{X}_{L}}=\sqrt{{{220}^{2}}-{{50}^{2}}}=214.24\Omega $
Since the connection is in series, the current will be the same 1A through the inductor and now we also have the inductive reactance. We could apply Ohm’s law here to get,
${{V}_{i}}=I{{X}_{L}}$
$\therefore {{V}_{i}}=1\times 214.24\approx 214V$
Therefore, we found the potential difference across the inductor to be 214V. Hence, option C is the correct answer.
Note: Impedance simply is the net opposition offered to the current in a particular circuit. The difference between impedance and resistance is that the impedance takes the effect of inductor and capacitor into account. The SI unit of impedance is also the same as resistance –ohms.
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