
A coil has a resistance of $30ohm$ and inductive reactance of $20ohm$ at $50Hz$. If an AC source of $200$ Volt, $100 Hz$, is connected across the coil, the current in the coil will be,
A. $4.0A$
B. $8.0A$
C. $\dfrac{20}{\sqrt{13}}A$
D. $2.0A$
Answer
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Hint: The inductive reactance of an inductor depends directly upon the frequency of an AC source while the capacitive reactance depends inversely on it. Resistance is independent of the frequency of the AC source. BY finding out the individual reactances and total impedance of the circuit, we can find out the current in the circuit by Ohm’s law.
Formula used:
${{X}_{L}}=2\pi fL$
where ${{X}_{L}}$ is the inductive reactance of an inductor, $f$ is the frequency of the AC source in the circuit and $L$ is the inductance of the inductor.
Impedance $Z$ of an AC circuit is given by,
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}$
where $R$ is the resistance in the circuit, ${{X}_{L}}$ is the inductive reactance and ${{X}_{C}}$ is the capacitive reactance.
According to Ohm’s law the current $\left( I \right)$ flowing in a circuit depends directly upon the potential drop $\left( V \right)$ across the circuit and inversely dependent upon the impedance $\left( Z \right)$ of the circuit. That is,
$I=\dfrac{V}{Z}$
Complete step by step answer:
We will solve this problem by first finding out the individual reactances and resistances of the elements in the circuit with the help of the information given. Then we will find the total impedance of the circuit and use Ohm’s law to find out the current flowing in the circuit due to the voltage provided by the AC source.
Hence, let us analyze the question.
We are given the inductive reactance for $50Hz$ but the source is $100Hz$. Hence, we need to find the inductive reactance for $100Hz$.
Now,
${{X}_{L}}=2\pi fL$ --(1)
where ${{X}_{L}}$ is the inductive reactance of an inductor, $f$ is the frequency of the AC source in the circuit and $L$ is the inductance of the inductor.
Hence, using (1), we see that for an inductor
${{X}_{L}}\propto f$
Hence, inductive reactance for $100Hz$ will be $\dfrac{\left( 100 \right)}{50}=2$ times the inductive reactance for $50Hz$.
Hence, by using the information in the question, the required inductive reactance will be
${{X}_{L}}=2\times 20\Omega =40\Omega $ --(2)
Now, since, there are no capacitors in the circuit, the capacitive reactance will be
zero.
$\therefore {{X}_{C}}=0$ --(3)
Resistance of the coil is independent of the frequency of the AC source.
Given resistance of the coil is $R=30\Omega $ --(4)
Impedance $Z$ of an AC circuit is given by,
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}$ --(5)
where $R$ is the resistance in the circuit, ${{X}_{L}}$ is the inductive reactance and ${{X}_{C}}$ is the capacitive reactance.
Using (2), (3) and (4) in (5), we get,
$Z=\sqrt{{{\left( 30 \right)}^{2}}+{{\left( 40-0 \right)}^{2}}}=\sqrt{900+1600}=\sqrt{2500}=50\Omega $ --(6)
According to Ohm’s law the current $\left( I \right)$ flowing in a circuit depends directly upon the potential drop $\left( V \right)$ across the circuit and inversely dependent upon the impedance $\left( Z \right)$ of the circuit. That is,
$I=\dfrac{V}{Z}$ --(7)
According to the question, voltage of the AC source is
$V=200Volt$ --(8)
Hence, putting (6) and (8) in (7), we get the current $I$ in the coil as
$I=\dfrac{200}{50}=4.0A$
Hence, the current in the coil is $4.0A$.
Therefore, the correct option is A) $4.0A$.
Note: Students can overlook the fact that the reactance of the coil in the question is given for a different frequency than that of the source and can proceed with the same values directly for the question. This will lead to a wrong answer. Students should always be careful that they have found the reactances for the same frequency as that of the AC source since, as seen above the reactances change with changing frequency of the AC source.
Formula used:
${{X}_{L}}=2\pi fL$
where ${{X}_{L}}$ is the inductive reactance of an inductor, $f$ is the frequency of the AC source in the circuit and $L$ is the inductance of the inductor.
Impedance $Z$ of an AC circuit is given by,
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}$
where $R$ is the resistance in the circuit, ${{X}_{L}}$ is the inductive reactance and ${{X}_{C}}$ is the capacitive reactance.
According to Ohm’s law the current $\left( I \right)$ flowing in a circuit depends directly upon the potential drop $\left( V \right)$ across the circuit and inversely dependent upon the impedance $\left( Z \right)$ of the circuit. That is,
$I=\dfrac{V}{Z}$
Complete step by step answer:
We will solve this problem by first finding out the individual reactances and resistances of the elements in the circuit with the help of the information given. Then we will find the total impedance of the circuit and use Ohm’s law to find out the current flowing in the circuit due to the voltage provided by the AC source.
Hence, let us analyze the question.
We are given the inductive reactance for $50Hz$ but the source is $100Hz$. Hence, we need to find the inductive reactance for $100Hz$.
Now,
${{X}_{L}}=2\pi fL$ --(1)
where ${{X}_{L}}$ is the inductive reactance of an inductor, $f$ is the frequency of the AC source in the circuit and $L$ is the inductance of the inductor.
Hence, using (1), we see that for an inductor
${{X}_{L}}\propto f$
Hence, inductive reactance for $100Hz$ will be $\dfrac{\left( 100 \right)}{50}=2$ times the inductive reactance for $50Hz$.
Hence, by using the information in the question, the required inductive reactance will be
${{X}_{L}}=2\times 20\Omega =40\Omega $ --(2)
Now, since, there are no capacitors in the circuit, the capacitive reactance will be
zero.
$\therefore {{X}_{C}}=0$ --(3)
Resistance of the coil is independent of the frequency of the AC source.
Given resistance of the coil is $R=30\Omega $ --(4)
Impedance $Z$ of an AC circuit is given by,
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}$ --(5)
where $R$ is the resistance in the circuit, ${{X}_{L}}$ is the inductive reactance and ${{X}_{C}}$ is the capacitive reactance.
Using (2), (3) and (4) in (5), we get,
$Z=\sqrt{{{\left( 30 \right)}^{2}}+{{\left( 40-0 \right)}^{2}}}=\sqrt{900+1600}=\sqrt{2500}=50\Omega $ --(6)
According to Ohm’s law the current $\left( I \right)$ flowing in a circuit depends directly upon the potential drop $\left( V \right)$ across the circuit and inversely dependent upon the impedance $\left( Z \right)$ of the circuit. That is,
$I=\dfrac{V}{Z}$ --(7)
According to the question, voltage of the AC source is
$V=200Volt$ --(8)
Hence, putting (6) and (8) in (7), we get the current $I$ in the coil as
$I=\dfrac{200}{50}=4.0A$
Hence, the current in the coil is $4.0A$.
Therefore, the correct option is A) $4.0A$.
Note: Students can overlook the fact that the reactance of the coil in the question is given for a different frequency than that of the source and can proceed with the same values directly for the question. This will lead to a wrong answer. Students should always be careful that they have found the reactances for the same frequency as that of the AC source since, as seen above the reactances change with changing frequency of the AC source.
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