155126
In LCR circuit the inductance is changed from $L$ to $9 \mathrm{~L}$. For same resonant frequency the capacitance should be changed from $\mathrm{C}$ to
1 $9 \mathrm{C}$
2 $\frac{\mathrm{C}}{9}$
3 $\frac{\mathrm{C}}{3}$
4 $3 \mathrm{C}$
Explanation:
B Given that, In LCR circuit inductance is change from $L$ to $9 \mathrm{~L}$ for same resonant frequency, $\omega=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}_{1} \quad\left(\because \mathrm{C}_{1}=\mathrm{C}\right)$ $\mathrm{C}_{2}=\frac{\mathrm{C}}{9}$
MHT-CET 2020
Alternating Current
155127
In an LCR circuit, inductive reactance is $30 \Omega$ and capacitive reactance $30 \Omega$. The resistance was found to be $40 \Omega$. The probable impedance of the combination is
1 $40 \Omega$
2 $60 \Omega$
3 $100 \Omega$
4 $20 \Omega$
Explanation:
A Given that, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=30 \Omega$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=30 \Omega$ Resistance $(\mathrm{R})=40 \Omega$ In LCR circuit impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(40)^{2}+(30-30)^{2}}$ $Z=\sqrt{(40)^{2}}$ $Z=40 \Omega$
MHT-CET 2020
Alternating Current
155130
The resonant frequency of a series LCR circuit is ' $f$ '. The circuit is now connected to the sinusoidally alternating e.m.f. of frequency ' $2 f$ '. The new reactance $X_{L}^{\prime}$ and $X_{C}^{\prime}$ are related as
A We know that, $\left(\mathrm{X}_{\mathrm{L}}\right)=2 \pi \mathrm{fL}, \mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ At resonant, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ When frequency is $2 \mathrm{f}$ then, $\mathrm{X}_{\mathrm{L}}^{\prime}=4 \pi \mathrm{fL} \quad \text { and } \quad \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{4 \pi \mathrm{fC}}$ So, $\quad X_{L}^{\prime}=2 X_{L}$ and $X_{C}^{\prime}=\frac{X_{C}}{2}$ Then, $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{\mathrm{X}_{\mathrm{C}} / 2}{2 \mathrm{X}_{\mathrm{L}}}=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{X}_{\mathrm{L}}} \times \frac{1}{4}$ $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{1}{4}$ $\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{4}$
MHT-CET 2020
Alternating Current
155133
An alternating e.m.f. of $0.2 \mathrm{~V}$ is applied across an LCR series circuit having $R=4 \Omega, C=80 \mu \mathrm{F}$ and $L=200 \mathrm{mH}$. At resonance the voltage drop across the inductor is
155126
In LCR circuit the inductance is changed from $L$ to $9 \mathrm{~L}$. For same resonant frequency the capacitance should be changed from $\mathrm{C}$ to
1 $9 \mathrm{C}$
2 $\frac{\mathrm{C}}{9}$
3 $\frac{\mathrm{C}}{3}$
4 $3 \mathrm{C}$
Explanation:
B Given that, In LCR circuit inductance is change from $L$ to $9 \mathrm{~L}$ for same resonant frequency, $\omega=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}_{1} \quad\left(\because \mathrm{C}_{1}=\mathrm{C}\right)$ $\mathrm{C}_{2}=\frac{\mathrm{C}}{9}$
MHT-CET 2020
Alternating Current
155127
In an LCR circuit, inductive reactance is $30 \Omega$ and capacitive reactance $30 \Omega$. The resistance was found to be $40 \Omega$. The probable impedance of the combination is
1 $40 \Omega$
2 $60 \Omega$
3 $100 \Omega$
4 $20 \Omega$
Explanation:
A Given that, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=30 \Omega$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=30 \Omega$ Resistance $(\mathrm{R})=40 \Omega$ In LCR circuit impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(40)^{2}+(30-30)^{2}}$ $Z=\sqrt{(40)^{2}}$ $Z=40 \Omega$
MHT-CET 2020
Alternating Current
155130
The resonant frequency of a series LCR circuit is ' $f$ '. The circuit is now connected to the sinusoidally alternating e.m.f. of frequency ' $2 f$ '. The new reactance $X_{L}^{\prime}$ and $X_{C}^{\prime}$ are related as
A We know that, $\left(\mathrm{X}_{\mathrm{L}}\right)=2 \pi \mathrm{fL}, \mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ At resonant, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ When frequency is $2 \mathrm{f}$ then, $\mathrm{X}_{\mathrm{L}}^{\prime}=4 \pi \mathrm{fL} \quad \text { and } \quad \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{4 \pi \mathrm{fC}}$ So, $\quad X_{L}^{\prime}=2 X_{L}$ and $X_{C}^{\prime}=\frac{X_{C}}{2}$ Then, $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{\mathrm{X}_{\mathrm{C}} / 2}{2 \mathrm{X}_{\mathrm{L}}}=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{X}_{\mathrm{L}}} \times \frac{1}{4}$ $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{1}{4}$ $\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{4}$
MHT-CET 2020
Alternating Current
155133
An alternating e.m.f. of $0.2 \mathrm{~V}$ is applied across an LCR series circuit having $R=4 \Omega, C=80 \mu \mathrm{F}$ and $L=200 \mathrm{mH}$. At resonance the voltage drop across the inductor is
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Alternating Current
155126
In LCR circuit the inductance is changed from $L$ to $9 \mathrm{~L}$. For same resonant frequency the capacitance should be changed from $\mathrm{C}$ to
1 $9 \mathrm{C}$
2 $\frac{\mathrm{C}}{9}$
3 $\frac{\mathrm{C}}{3}$
4 $3 \mathrm{C}$
Explanation:
B Given that, In LCR circuit inductance is change from $L$ to $9 \mathrm{~L}$ for same resonant frequency, $\omega=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}_{1} \quad\left(\because \mathrm{C}_{1}=\mathrm{C}\right)$ $\mathrm{C}_{2}=\frac{\mathrm{C}}{9}$
MHT-CET 2020
Alternating Current
155127
In an LCR circuit, inductive reactance is $30 \Omega$ and capacitive reactance $30 \Omega$. The resistance was found to be $40 \Omega$. The probable impedance of the combination is
1 $40 \Omega$
2 $60 \Omega$
3 $100 \Omega$
4 $20 \Omega$
Explanation:
A Given that, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=30 \Omega$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=30 \Omega$ Resistance $(\mathrm{R})=40 \Omega$ In LCR circuit impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(40)^{2}+(30-30)^{2}}$ $Z=\sqrt{(40)^{2}}$ $Z=40 \Omega$
MHT-CET 2020
Alternating Current
155130
The resonant frequency of a series LCR circuit is ' $f$ '. The circuit is now connected to the sinusoidally alternating e.m.f. of frequency ' $2 f$ '. The new reactance $X_{L}^{\prime}$ and $X_{C}^{\prime}$ are related as
A We know that, $\left(\mathrm{X}_{\mathrm{L}}\right)=2 \pi \mathrm{fL}, \mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ At resonant, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ When frequency is $2 \mathrm{f}$ then, $\mathrm{X}_{\mathrm{L}}^{\prime}=4 \pi \mathrm{fL} \quad \text { and } \quad \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{4 \pi \mathrm{fC}}$ So, $\quad X_{L}^{\prime}=2 X_{L}$ and $X_{C}^{\prime}=\frac{X_{C}}{2}$ Then, $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{\mathrm{X}_{\mathrm{C}} / 2}{2 \mathrm{X}_{\mathrm{L}}}=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{X}_{\mathrm{L}}} \times \frac{1}{4}$ $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{1}{4}$ $\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{4}$
MHT-CET 2020
Alternating Current
155133
An alternating e.m.f. of $0.2 \mathrm{~V}$ is applied across an LCR series circuit having $R=4 \Omega, C=80 \mu \mathrm{F}$ and $L=200 \mathrm{mH}$. At resonance the voltage drop across the inductor is
155126
In LCR circuit the inductance is changed from $L$ to $9 \mathrm{~L}$. For same resonant frequency the capacitance should be changed from $\mathrm{C}$ to
1 $9 \mathrm{C}$
2 $\frac{\mathrm{C}}{9}$
3 $\frac{\mathrm{C}}{3}$
4 $3 \mathrm{C}$
Explanation:
B Given that, In LCR circuit inductance is change from $L$ to $9 \mathrm{~L}$ for same resonant frequency, $\omega=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}_{1} \quad\left(\because \mathrm{C}_{1}=\mathrm{C}\right)$ $\mathrm{C}_{2}=\frac{\mathrm{C}}{9}$
MHT-CET 2020
Alternating Current
155127
In an LCR circuit, inductive reactance is $30 \Omega$ and capacitive reactance $30 \Omega$. The resistance was found to be $40 \Omega$. The probable impedance of the combination is
1 $40 \Omega$
2 $60 \Omega$
3 $100 \Omega$
4 $20 \Omega$
Explanation:
A Given that, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=30 \Omega$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=30 \Omega$ Resistance $(\mathrm{R})=40 \Omega$ In LCR circuit impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(40)^{2}+(30-30)^{2}}$ $Z=\sqrt{(40)^{2}}$ $Z=40 \Omega$
MHT-CET 2020
Alternating Current
155130
The resonant frequency of a series LCR circuit is ' $f$ '. The circuit is now connected to the sinusoidally alternating e.m.f. of frequency ' $2 f$ '. The new reactance $X_{L}^{\prime}$ and $X_{C}^{\prime}$ are related as
A We know that, $\left(\mathrm{X}_{\mathrm{L}}\right)=2 \pi \mathrm{fL}, \mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ At resonant, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ When frequency is $2 \mathrm{f}$ then, $\mathrm{X}_{\mathrm{L}}^{\prime}=4 \pi \mathrm{fL} \quad \text { and } \quad \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{4 \pi \mathrm{fC}}$ So, $\quad X_{L}^{\prime}=2 X_{L}$ and $X_{C}^{\prime}=\frac{X_{C}}{2}$ Then, $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{\mathrm{X}_{\mathrm{C}} / 2}{2 \mathrm{X}_{\mathrm{L}}}=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{X}_{\mathrm{L}}} \times \frac{1}{4}$ $\frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{L}}^{\prime}}=\frac{1}{4}$ $\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{4}$
MHT-CET 2020
Alternating Current
155133
An alternating e.m.f. of $0.2 \mathrm{~V}$ is applied across an LCR series circuit having $R=4 \Omega, C=80 \mu \mathrm{F}$ and $L=200 \mathrm{mH}$. At resonance the voltage drop across the inductor is
