
An inductor 20mh, a capacitor $100\mu F$ and a resistor $50\Omega $ are connected in series across a source of emf, \[V=10\sin 314t\]. The power loss in the circuit is?
\[\begin{array}{*{35}{l}}
\text{A}\text{. }2.74W \\
\text{B}\text{. }0.79W \\
\text{C}\text{. }1.1W \\
\text{D}\text{. }0.43W \\
\end{array}\]
Answer
581.1k+ views
Hint: This question can be solved using the concept of ‘A.C. thought a series combination of pure inductor, capacitor and resistor circuit’. Resistor, inductor and capacitor are offering opposition to the series combination circuit with alternating e.m.f. Connect to circuit. Use formula of power loss derived for LCR alternating circuit and calculate value of impedance denoted by Z.
Complete step by step answer:
In this question, it is given that the inductor has 20mh and capacitor has capacitance $100\mu F$ and resistor has $50\Omega $ resistance.
I.e.\[L=20mH,C=100\mu F,R=50\Omega \]
We know that voltage in alternating circuit (having frequency) is in the form $V={{V}_{0}}\sin (\omega t)$
And we have $V=10\sin (314t)$
Therefore, ${{V}_{0}}=10,\omega =314$
Inductance of inductor is given by ${{X}_{L}}=\omega L=314\times 20\times {{10}^{-3}}=6.28\Omega $
Capacitance of capacitor is given by ${{X}_{C}}=\dfrac{1}{\omega C}=\dfrac{1}{314\times 100\times {{10}^{-6}}}=0.00318\times {{10}^{4}}=31.8\Omega $
We know that power loss is given by $P=\dfrac{{{V}_{0}}^{2}R}{2{{Z}^{2}}}$
We know value of all the quantity in power formula except Z which is given by $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$
Put value in above equation from given data, we get
\[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=\sqrt{{{50}^{2}}+{{(6.28-31.8)}^{2}}}=\sqrt{2500+65.27}=\sqrt{31.51}=56.1\]
Therefore power loss is given by $P=\dfrac{{{V}_{0}}^{2}R}{2{{Z}^{2}}}=\dfrac{{{(10)}^{2}}50}{2{{(56.1)}^{2}}}=0.79W$
The power loss in the circuit is $0.79W$. Therefore the correct option is (B).
Note:
V in question gives r.m.s value of voltage across the circuit. Quantity Z is impedance of circuit which represents effective opposition offered by the L, C and R connected in series to the flow of a.c. Current. Impedance of circuit can also be defined as ratio of r.m.s voltage I.e. E.m.f to r.m.s value of current. Unit of impedance is ohm white unit of power is watt. If a capacitor is absent in an LCR circuit then circuits are called a LR circuit.
Complete step by step answer:
In this question, it is given that the inductor has 20mh and capacitor has capacitance $100\mu F$ and resistor has $50\Omega $ resistance.
I.e.\[L=20mH,C=100\mu F,R=50\Omega \]
We know that voltage in alternating circuit (having frequency) is in the form $V={{V}_{0}}\sin (\omega t)$
And we have $V=10\sin (314t)$
Therefore, ${{V}_{0}}=10,\omega =314$
Inductance of inductor is given by ${{X}_{L}}=\omega L=314\times 20\times {{10}^{-3}}=6.28\Omega $
Capacitance of capacitor is given by ${{X}_{C}}=\dfrac{1}{\omega C}=\dfrac{1}{314\times 100\times {{10}^{-6}}}=0.00318\times {{10}^{4}}=31.8\Omega $
We know that power loss is given by $P=\dfrac{{{V}_{0}}^{2}R}{2{{Z}^{2}}}$
We know value of all the quantity in power formula except Z which is given by $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$
Put value in above equation from given data, we get
\[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=\sqrt{{{50}^{2}}+{{(6.28-31.8)}^{2}}}=\sqrt{2500+65.27}=\sqrt{31.51}=56.1\]
Therefore power loss is given by $P=\dfrac{{{V}_{0}}^{2}R}{2{{Z}^{2}}}=\dfrac{{{(10)}^{2}}50}{2{{(56.1)}^{2}}}=0.79W$
The power loss in the circuit is $0.79W$. Therefore the correct option is (B).
Note:
V in question gives r.m.s value of voltage across the circuit. Quantity Z is impedance of circuit which represents effective opposition offered by the L, C and R connected in series to the flow of a.c. Current. Impedance of circuit can also be defined as ratio of r.m.s voltage I.e. E.m.f to r.m.s value of current. Unit of impedance is ohm white unit of power is watt. If a capacitor is absent in an LCR circuit then circuits are called a LR circuit.
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