
To reduce current in an ac circuit, a choke coil is preferred over a resistance. This is so because
(A) choke coil is easier to design
(B) choke coil is cheaper
(C) choke coil consumes much less power
(D) the eddy currents produced in a choke coil quickly reduce the current in the circuit
Answer
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Hint:Here, you are asked the reason for taking choke coil over resistance in order to reduce current in an ac circuit. So, to proceed, you need to consider the properties of a choke coil. A choke coil is simply a coil with large inductance and small resistance. Analyze the choke coil when the current passes through it.
Complete answer:
Choke coil is a coil with large inductance and small resistance. It is used in an ac circuit in order to reduce or control the current in the circuit. Let us consider a circuit which includes a voltage source which supplies a voltage of $V = {V_0}\sin \omega t$, a resistor of resistance $R$ and a component $Y$ which can be a choke coil or a resistance. First, let $Y$ be a choke coil.
As discussed before, choke is nothing but an inductor, so, the resulting circuit will be an $LR$ circuit with impedance $Z = \sqrt {{R^2} + {\omega ^2}{L^2}} $. The peak current in the circuit will be ${I_0} = \dfrac{{{V_0}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}$ and the rms current will be ${I_{rms}} = \dfrac{{\dfrac{{{V_0}}}{{\sqrt 2 }}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}\dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}$. The rms voltage across the resistor will be \[\dfrac{R}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}{V_{rms}}\]. The voltage across the resistor is reduced by a factor of \[\dfrac{R}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}\].
Now, consider the case where $Y$ is a resistor. This much reduction factor will not be obtained in the present case if you compare it with a choke coil. As a result, less power is consumed.Therefore, to reduce current in an ac circuit, a choke coil is preferred over a resistance because the choke coil consumes much less power.
Hence, option C is correct.
Note:You need to keep in mind that the choke coil reduces the voltage across the resistor which in turn reduces the power consumed. In case if a resistor is used instead of an inductor, due to heat dissipation according to Joule heating effect in resistances.
Complete answer:
Choke coil is a coil with large inductance and small resistance. It is used in an ac circuit in order to reduce or control the current in the circuit. Let us consider a circuit which includes a voltage source which supplies a voltage of $V = {V_0}\sin \omega t$, a resistor of resistance $R$ and a component $Y$ which can be a choke coil or a resistance. First, let $Y$ be a choke coil.
As discussed before, choke is nothing but an inductor, so, the resulting circuit will be an $LR$ circuit with impedance $Z = \sqrt {{R^2} + {\omega ^2}{L^2}} $. The peak current in the circuit will be ${I_0} = \dfrac{{{V_0}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}$ and the rms current will be ${I_{rms}} = \dfrac{{\dfrac{{{V_0}}}{{\sqrt 2 }}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}\dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}$. The rms voltage across the resistor will be \[\dfrac{R}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}{V_{rms}}\]. The voltage across the resistor is reduced by a factor of \[\dfrac{R}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }}\].
Now, consider the case where $Y$ is a resistor. This much reduction factor will not be obtained in the present case if you compare it with a choke coil. As a result, less power is consumed.Therefore, to reduce current in an ac circuit, a choke coil is preferred over a resistance because the choke coil consumes much less power.
Hence, option C is correct.
Note:You need to keep in mind that the choke coil reduces the voltage across the resistor which in turn reduces the power consumed. In case if a resistor is used instead of an inductor, due to heat dissipation according to Joule heating effect in resistances.
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