155059
Two inductors each of inductance $L$ are connected in parallel. One more inductor of value $5 \mathrm{mH}$ is connected in series of this configuration then the effective inductance is $15 \mathrm{mH}$. The value of $\mathrm{L}$ is mH.
1 10
2 5.0
3 2.5
4 20
Explanation:
D Equivalent inductance in parallel connection, $\frac{1}{\mathrm{~L}^{\prime}}=\frac{1}{\mathrm{~L}_{1}}+\frac{1}{\mathrm{~L}_{2}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{1}{\mathrm{~L}}+\frac{1}{\mathrm{~L}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{2}{\mathrm{~L}}$ $\mathrm{~L}^{\prime} =\frac{\mathrm{L}}{2}$ This combination connected in series with $\mathrm{L}_{3}=5 \mathrm{mH}$ $1_{\text {eq }}=L^{\prime}+L_{3}$ $15 \mathrm{mH}=\frac{\mathrm{L}}{2}+5 \Rightarrow 30=\mathrm{L}+10$ $\mathrm{~L}=20 \mathrm{mH}$
GUJCET 2018
Alternating Current
155062
A $30 \mu \mathrm{F}$ capacitor is connected to a $150 \mathrm{~V}, 60$ $\mathrm{Hz}$ AC supply the rms value of current is the circuit is
1 $17 \mathrm{~A}$
2 $1.7 \mathrm{~A}$
3 $1.7 \mathrm{~mA}$
4 $1.7 \mu \mathrm{A}$
Explanation:
B Given that, Capacitance $(\mathrm{C})=30 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$ rms voltage $\left(\mathrm{V}_{\mathrm{rms}}\right)=150 \mathrm{~V}$, Frequency $(\mathrm{f})=60 \mathrm{~Hz}$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=\frac{1}{\omega \mathrm{C}}$ $=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \times 3.14 \times 60 \times 30 \times 10^{-6}}$ The rms value of current is, $I_{\text {rms }}= \frac{V_{\text {rms }}}{X_{C}}$ $=150 \times 3.14 \times 2 \times 60 \times 30 \times 10^{-6}=1.7 \mathrm{~A}$
JCECE-2018
Alternating Current
155063
An oscillator circuit contains an inductor 0.05 $H$ and a capacitor of capacity $80 \mu \mathrm{F}$. When the maximum voltage across the capacitor is $200 \mathrm{~V}$, the maximum current (in amperes) in the circuit is
1 2
2 4
3 8
4 10
5 16
Explanation:
C Given that, Inductance $(\mathrm{L})=0.05 \mathrm{H}$, Capacitance $(\mathrm{C})=80 \mu \mathrm{F}$, Maximum Voltage $\left(\mathrm{V}_{\max }\right)=\mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$ $\mathrm{V}(\mathrm{t})=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$ $\because \mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{CV}_{\mathrm{m}} \omega \cos \omega \mathrm{t}$ $\therefore \mathrm{\omega}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \mathrm{i}=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \cos \omega \mathrm{t}$ For maximum current, $\cos \omega \mathrm{t}=1$ So, Max. current $\left(\mathrm{i}_{\mathrm{m}}\right)=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}=200 \times \sqrt{\frac{80 \times 10^{-6}}{0.05}}=8 \mathrm{~A}$
Kerala CEE -2018
Alternating Current
155064
A coil of inductive reactance $1 / \sqrt{3} \Omega$ and resistance $1 \Omega$ is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply. The time lag between maximum voltage and current is :
1 $\frac{1}{200} \mathrm{~s}$
2 $\frac{1}{300} \mathrm{~s}$
3 $\frac{1}{500} \mathrm{~s}$
4 $\frac{1}{600} \mathrm{~s}$
Explanation:
D Given, Ans: a : Given that, Potential drop across the capacitor $=5 \mathrm{~V}$ Potential drop across the resistor $=12 \mathrm{~V}$ Since, $\quad V=\sqrt{V_{R}^{2}+V_{C}^{2}}$ Applied voltage $\left(\mathrm{V}_{\mathrm{ap}}\right)=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}$ $=\sqrt{(5)^{2}+(12)^{2}}$ $\mathrm{~V}_{\text {ap }} =\sqrt{169}$ $\mathrm{~V}_{\mathrm{ap}} =13 \mathrm{~V}$ Hence, the applied voltage $\mathrm{V}_{\text {ap }}=13 \mathrm{~V}$.
Manipal UGET-2009
Alternating Current
155066
A L-R circuit consists of an inductance of $8 \mathrm{mH}$ and a resistance of $4 \Omega$. The time constant of the circuit is
1 $2 \mathrm{~ms}$
2 $12 \mathrm{~ms}$
3 $32 \mathrm{~ms}$
4 $500 \mathrm{~s}$
Explanation:
A Given that, Inductance $(\mathrm{L})=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=4 \Omega$ In L-R circuit, Time constant $(\mathrm{t})=\frac{\mathrm{L}}{\mathrm{R}}=\frac{8 \times 10^{-3}}{4}$ $=2 \times 10^{-3} \text { second }=2 \mathrm{~ms}$
155059
Two inductors each of inductance $L$ are connected in parallel. One more inductor of value $5 \mathrm{mH}$ is connected in series of this configuration then the effective inductance is $15 \mathrm{mH}$. The value of $\mathrm{L}$ is mH.
1 10
2 5.0
3 2.5
4 20
Explanation:
D Equivalent inductance in parallel connection, $\frac{1}{\mathrm{~L}^{\prime}}=\frac{1}{\mathrm{~L}_{1}}+\frac{1}{\mathrm{~L}_{2}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{1}{\mathrm{~L}}+\frac{1}{\mathrm{~L}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{2}{\mathrm{~L}}$ $\mathrm{~L}^{\prime} =\frac{\mathrm{L}}{2}$ This combination connected in series with $\mathrm{L}_{3}=5 \mathrm{mH}$ $1_{\text {eq }}=L^{\prime}+L_{3}$ $15 \mathrm{mH}=\frac{\mathrm{L}}{2}+5 \Rightarrow 30=\mathrm{L}+10$ $\mathrm{~L}=20 \mathrm{mH}$
GUJCET 2018
Alternating Current
155062
A $30 \mu \mathrm{F}$ capacitor is connected to a $150 \mathrm{~V}, 60$ $\mathrm{Hz}$ AC supply the rms value of current is the circuit is
1 $17 \mathrm{~A}$
2 $1.7 \mathrm{~A}$
3 $1.7 \mathrm{~mA}$
4 $1.7 \mu \mathrm{A}$
Explanation:
B Given that, Capacitance $(\mathrm{C})=30 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$ rms voltage $\left(\mathrm{V}_{\mathrm{rms}}\right)=150 \mathrm{~V}$, Frequency $(\mathrm{f})=60 \mathrm{~Hz}$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=\frac{1}{\omega \mathrm{C}}$ $=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \times 3.14 \times 60 \times 30 \times 10^{-6}}$ The rms value of current is, $I_{\text {rms }}= \frac{V_{\text {rms }}}{X_{C}}$ $=150 \times 3.14 \times 2 \times 60 \times 30 \times 10^{-6}=1.7 \mathrm{~A}$
JCECE-2018
Alternating Current
155063
An oscillator circuit contains an inductor 0.05 $H$ and a capacitor of capacity $80 \mu \mathrm{F}$. When the maximum voltage across the capacitor is $200 \mathrm{~V}$, the maximum current (in amperes) in the circuit is
1 2
2 4
3 8
4 10
5 16
Explanation:
C Given that, Inductance $(\mathrm{L})=0.05 \mathrm{H}$, Capacitance $(\mathrm{C})=80 \mu \mathrm{F}$, Maximum Voltage $\left(\mathrm{V}_{\max }\right)=\mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$ $\mathrm{V}(\mathrm{t})=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$ $\because \mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{CV}_{\mathrm{m}} \omega \cos \omega \mathrm{t}$ $\therefore \mathrm{\omega}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \mathrm{i}=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \cos \omega \mathrm{t}$ For maximum current, $\cos \omega \mathrm{t}=1$ So, Max. current $\left(\mathrm{i}_{\mathrm{m}}\right)=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}=200 \times \sqrt{\frac{80 \times 10^{-6}}{0.05}}=8 \mathrm{~A}$
Kerala CEE -2018
Alternating Current
155064
A coil of inductive reactance $1 / \sqrt{3} \Omega$ and resistance $1 \Omega$ is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply. The time lag between maximum voltage and current is :
1 $\frac{1}{200} \mathrm{~s}$
2 $\frac{1}{300} \mathrm{~s}$
3 $\frac{1}{500} \mathrm{~s}$
4 $\frac{1}{600} \mathrm{~s}$
Explanation:
D Given, Ans: a : Given that, Potential drop across the capacitor $=5 \mathrm{~V}$ Potential drop across the resistor $=12 \mathrm{~V}$ Since, $\quad V=\sqrt{V_{R}^{2}+V_{C}^{2}}$ Applied voltage $\left(\mathrm{V}_{\mathrm{ap}}\right)=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}$ $=\sqrt{(5)^{2}+(12)^{2}}$ $\mathrm{~V}_{\text {ap }} =\sqrt{169}$ $\mathrm{~V}_{\mathrm{ap}} =13 \mathrm{~V}$ Hence, the applied voltage $\mathrm{V}_{\text {ap }}=13 \mathrm{~V}$.
Manipal UGET-2009
Alternating Current
155066
A L-R circuit consists of an inductance of $8 \mathrm{mH}$ and a resistance of $4 \Omega$. The time constant of the circuit is
1 $2 \mathrm{~ms}$
2 $12 \mathrm{~ms}$
3 $32 \mathrm{~ms}$
4 $500 \mathrm{~s}$
Explanation:
A Given that, Inductance $(\mathrm{L})=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=4 \Omega$ In L-R circuit, Time constant $(\mathrm{t})=\frac{\mathrm{L}}{\mathrm{R}}=\frac{8 \times 10^{-3}}{4}$ $=2 \times 10^{-3} \text { second }=2 \mathrm{~ms}$
155059
Two inductors each of inductance $L$ are connected in parallel. One more inductor of value $5 \mathrm{mH}$ is connected in series of this configuration then the effective inductance is $15 \mathrm{mH}$. The value of $\mathrm{L}$ is mH.
1 10
2 5.0
3 2.5
4 20
Explanation:
D Equivalent inductance in parallel connection, $\frac{1}{\mathrm{~L}^{\prime}}=\frac{1}{\mathrm{~L}_{1}}+\frac{1}{\mathrm{~L}_{2}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{1}{\mathrm{~L}}+\frac{1}{\mathrm{~L}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{2}{\mathrm{~L}}$ $\mathrm{~L}^{\prime} =\frac{\mathrm{L}}{2}$ This combination connected in series with $\mathrm{L}_{3}=5 \mathrm{mH}$ $1_{\text {eq }}=L^{\prime}+L_{3}$ $15 \mathrm{mH}=\frac{\mathrm{L}}{2}+5 \Rightarrow 30=\mathrm{L}+10$ $\mathrm{~L}=20 \mathrm{mH}$
GUJCET 2018
Alternating Current
155062
A $30 \mu \mathrm{F}$ capacitor is connected to a $150 \mathrm{~V}, 60$ $\mathrm{Hz}$ AC supply the rms value of current is the circuit is
1 $17 \mathrm{~A}$
2 $1.7 \mathrm{~A}$
3 $1.7 \mathrm{~mA}$
4 $1.7 \mu \mathrm{A}$
Explanation:
B Given that, Capacitance $(\mathrm{C})=30 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$ rms voltage $\left(\mathrm{V}_{\mathrm{rms}}\right)=150 \mathrm{~V}$, Frequency $(\mathrm{f})=60 \mathrm{~Hz}$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=\frac{1}{\omega \mathrm{C}}$ $=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \times 3.14 \times 60 \times 30 \times 10^{-6}}$ The rms value of current is, $I_{\text {rms }}= \frac{V_{\text {rms }}}{X_{C}}$ $=150 \times 3.14 \times 2 \times 60 \times 30 \times 10^{-6}=1.7 \mathrm{~A}$
JCECE-2018
Alternating Current
155063
An oscillator circuit contains an inductor 0.05 $H$ and a capacitor of capacity $80 \mu \mathrm{F}$. When the maximum voltage across the capacitor is $200 \mathrm{~V}$, the maximum current (in amperes) in the circuit is
1 2
2 4
3 8
4 10
5 16
Explanation:
C Given that, Inductance $(\mathrm{L})=0.05 \mathrm{H}$, Capacitance $(\mathrm{C})=80 \mu \mathrm{F}$, Maximum Voltage $\left(\mathrm{V}_{\max }\right)=\mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$ $\mathrm{V}(\mathrm{t})=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$ $\because \mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{CV}_{\mathrm{m}} \omega \cos \omega \mathrm{t}$ $\therefore \mathrm{\omega}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \mathrm{i}=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \cos \omega \mathrm{t}$ For maximum current, $\cos \omega \mathrm{t}=1$ So, Max. current $\left(\mathrm{i}_{\mathrm{m}}\right)=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}=200 \times \sqrt{\frac{80 \times 10^{-6}}{0.05}}=8 \mathrm{~A}$
Kerala CEE -2018
Alternating Current
155064
A coil of inductive reactance $1 / \sqrt{3} \Omega$ and resistance $1 \Omega$ is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply. The time lag between maximum voltage and current is :
1 $\frac{1}{200} \mathrm{~s}$
2 $\frac{1}{300} \mathrm{~s}$
3 $\frac{1}{500} \mathrm{~s}$
4 $\frac{1}{600} \mathrm{~s}$
Explanation:
D Given, Ans: a : Given that, Potential drop across the capacitor $=5 \mathrm{~V}$ Potential drop across the resistor $=12 \mathrm{~V}$ Since, $\quad V=\sqrt{V_{R}^{2}+V_{C}^{2}}$ Applied voltage $\left(\mathrm{V}_{\mathrm{ap}}\right)=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}$ $=\sqrt{(5)^{2}+(12)^{2}}$ $\mathrm{~V}_{\text {ap }} =\sqrt{169}$ $\mathrm{~V}_{\mathrm{ap}} =13 \mathrm{~V}$ Hence, the applied voltage $\mathrm{V}_{\text {ap }}=13 \mathrm{~V}$.
Manipal UGET-2009
Alternating Current
155066
A L-R circuit consists of an inductance of $8 \mathrm{mH}$ and a resistance of $4 \Omega$. The time constant of the circuit is
1 $2 \mathrm{~ms}$
2 $12 \mathrm{~ms}$
3 $32 \mathrm{~ms}$
4 $500 \mathrm{~s}$
Explanation:
A Given that, Inductance $(\mathrm{L})=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=4 \Omega$ In L-R circuit, Time constant $(\mathrm{t})=\frac{\mathrm{L}}{\mathrm{R}}=\frac{8 \times 10^{-3}}{4}$ $=2 \times 10^{-3} \text { second }=2 \mathrm{~ms}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155059
Two inductors each of inductance $L$ are connected in parallel. One more inductor of value $5 \mathrm{mH}$ is connected in series of this configuration then the effective inductance is $15 \mathrm{mH}$. The value of $\mathrm{L}$ is mH.
1 10
2 5.0
3 2.5
4 20
Explanation:
D Equivalent inductance in parallel connection, $\frac{1}{\mathrm{~L}^{\prime}}=\frac{1}{\mathrm{~L}_{1}}+\frac{1}{\mathrm{~L}_{2}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{1}{\mathrm{~L}}+\frac{1}{\mathrm{~L}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{2}{\mathrm{~L}}$ $\mathrm{~L}^{\prime} =\frac{\mathrm{L}}{2}$ This combination connected in series with $\mathrm{L}_{3}=5 \mathrm{mH}$ $1_{\text {eq }}=L^{\prime}+L_{3}$ $15 \mathrm{mH}=\frac{\mathrm{L}}{2}+5 \Rightarrow 30=\mathrm{L}+10$ $\mathrm{~L}=20 \mathrm{mH}$
GUJCET 2018
Alternating Current
155062
A $30 \mu \mathrm{F}$ capacitor is connected to a $150 \mathrm{~V}, 60$ $\mathrm{Hz}$ AC supply the rms value of current is the circuit is
1 $17 \mathrm{~A}$
2 $1.7 \mathrm{~A}$
3 $1.7 \mathrm{~mA}$
4 $1.7 \mu \mathrm{A}$
Explanation:
B Given that, Capacitance $(\mathrm{C})=30 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$ rms voltage $\left(\mathrm{V}_{\mathrm{rms}}\right)=150 \mathrm{~V}$, Frequency $(\mathrm{f})=60 \mathrm{~Hz}$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=\frac{1}{\omega \mathrm{C}}$ $=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \times 3.14 \times 60 \times 30 \times 10^{-6}}$ The rms value of current is, $I_{\text {rms }}= \frac{V_{\text {rms }}}{X_{C}}$ $=150 \times 3.14 \times 2 \times 60 \times 30 \times 10^{-6}=1.7 \mathrm{~A}$
JCECE-2018
Alternating Current
155063
An oscillator circuit contains an inductor 0.05 $H$ and a capacitor of capacity $80 \mu \mathrm{F}$. When the maximum voltage across the capacitor is $200 \mathrm{~V}$, the maximum current (in amperes) in the circuit is
1 2
2 4
3 8
4 10
5 16
Explanation:
C Given that, Inductance $(\mathrm{L})=0.05 \mathrm{H}$, Capacitance $(\mathrm{C})=80 \mu \mathrm{F}$, Maximum Voltage $\left(\mathrm{V}_{\max }\right)=\mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$ $\mathrm{V}(\mathrm{t})=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$ $\because \mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{CV}_{\mathrm{m}} \omega \cos \omega \mathrm{t}$ $\therefore \mathrm{\omega}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \mathrm{i}=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \cos \omega \mathrm{t}$ For maximum current, $\cos \omega \mathrm{t}=1$ So, Max. current $\left(\mathrm{i}_{\mathrm{m}}\right)=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}=200 \times \sqrt{\frac{80 \times 10^{-6}}{0.05}}=8 \mathrm{~A}$
Kerala CEE -2018
Alternating Current
155064
A coil of inductive reactance $1 / \sqrt{3} \Omega$ and resistance $1 \Omega$ is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply. The time lag between maximum voltage and current is :
1 $\frac{1}{200} \mathrm{~s}$
2 $\frac{1}{300} \mathrm{~s}$
3 $\frac{1}{500} \mathrm{~s}$
4 $\frac{1}{600} \mathrm{~s}$
Explanation:
D Given, Ans: a : Given that, Potential drop across the capacitor $=5 \mathrm{~V}$ Potential drop across the resistor $=12 \mathrm{~V}$ Since, $\quad V=\sqrt{V_{R}^{2}+V_{C}^{2}}$ Applied voltage $\left(\mathrm{V}_{\mathrm{ap}}\right)=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}$ $=\sqrt{(5)^{2}+(12)^{2}}$ $\mathrm{~V}_{\text {ap }} =\sqrt{169}$ $\mathrm{~V}_{\mathrm{ap}} =13 \mathrm{~V}$ Hence, the applied voltage $\mathrm{V}_{\text {ap }}=13 \mathrm{~V}$.
Manipal UGET-2009
Alternating Current
155066
A L-R circuit consists of an inductance of $8 \mathrm{mH}$ and a resistance of $4 \Omega$. The time constant of the circuit is
1 $2 \mathrm{~ms}$
2 $12 \mathrm{~ms}$
3 $32 \mathrm{~ms}$
4 $500 \mathrm{~s}$
Explanation:
A Given that, Inductance $(\mathrm{L})=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=4 \Omega$ In L-R circuit, Time constant $(\mathrm{t})=\frac{\mathrm{L}}{\mathrm{R}}=\frac{8 \times 10^{-3}}{4}$ $=2 \times 10^{-3} \text { second }=2 \mathrm{~ms}$
155059
Two inductors each of inductance $L$ are connected in parallel. One more inductor of value $5 \mathrm{mH}$ is connected in series of this configuration then the effective inductance is $15 \mathrm{mH}$. The value of $\mathrm{L}$ is mH.
1 10
2 5.0
3 2.5
4 20
Explanation:
D Equivalent inductance in parallel connection, $\frac{1}{\mathrm{~L}^{\prime}}=\frac{1}{\mathrm{~L}_{1}}+\frac{1}{\mathrm{~L}_{2}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{1}{\mathrm{~L}}+\frac{1}{\mathrm{~L}}$ $\frac{1}{\mathrm{~L}^{\prime}} =\frac{2}{\mathrm{~L}}$ $\mathrm{~L}^{\prime} =\frac{\mathrm{L}}{2}$ This combination connected in series with $\mathrm{L}_{3}=5 \mathrm{mH}$ $1_{\text {eq }}=L^{\prime}+L_{3}$ $15 \mathrm{mH}=\frac{\mathrm{L}}{2}+5 \Rightarrow 30=\mathrm{L}+10$ $\mathrm{~L}=20 \mathrm{mH}$
GUJCET 2018
Alternating Current
155062
A $30 \mu \mathrm{F}$ capacitor is connected to a $150 \mathrm{~V}, 60$ $\mathrm{Hz}$ AC supply the rms value of current is the circuit is
1 $17 \mathrm{~A}$
2 $1.7 \mathrm{~A}$
3 $1.7 \mathrm{~mA}$
4 $1.7 \mu \mathrm{A}$
Explanation:
B Given that, Capacitance $(\mathrm{C})=30 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$ rms voltage $\left(\mathrm{V}_{\mathrm{rms}}\right)=150 \mathrm{~V}$, Frequency $(\mathrm{f})=60 \mathrm{~Hz}$ Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=\frac{1}{\omega \mathrm{C}}$ $=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \times 3.14 \times 60 \times 30 \times 10^{-6}}$ The rms value of current is, $I_{\text {rms }}= \frac{V_{\text {rms }}}{X_{C}}$ $=150 \times 3.14 \times 2 \times 60 \times 30 \times 10^{-6}=1.7 \mathrm{~A}$
JCECE-2018
Alternating Current
155063
An oscillator circuit contains an inductor 0.05 $H$ and a capacitor of capacity $80 \mu \mathrm{F}$. When the maximum voltage across the capacitor is $200 \mathrm{~V}$, the maximum current (in amperes) in the circuit is
1 2
2 4
3 8
4 10
5 16
Explanation:
C Given that, Inductance $(\mathrm{L})=0.05 \mathrm{H}$, Capacitance $(\mathrm{C})=80 \mu \mathrm{F}$, Maximum Voltage $\left(\mathrm{V}_{\max }\right)=\mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$ $\mathrm{V}(\mathrm{t})=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$ $\because \mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{CV}_{\mathrm{m}} \omega \cos \omega \mathrm{t}$ $\therefore \mathrm{\omega}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \mathrm{i}=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \cos \omega \mathrm{t}$ For maximum current, $\cos \omega \mathrm{t}=1$ So, Max. current $\left(\mathrm{i}_{\mathrm{m}}\right)=\mathrm{V}_{\mathrm{m}} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}=200 \times \sqrt{\frac{80 \times 10^{-6}}{0.05}}=8 \mathrm{~A}$
Kerala CEE -2018
Alternating Current
155064
A coil of inductive reactance $1 / \sqrt{3} \Omega$ and resistance $1 \Omega$ is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply. The time lag between maximum voltage and current is :
1 $\frac{1}{200} \mathrm{~s}$
2 $\frac{1}{300} \mathrm{~s}$
3 $\frac{1}{500} \mathrm{~s}$
4 $\frac{1}{600} \mathrm{~s}$
Explanation:
D Given, Ans: a : Given that, Potential drop across the capacitor $=5 \mathrm{~V}$ Potential drop across the resistor $=12 \mathrm{~V}$ Since, $\quad V=\sqrt{V_{R}^{2}+V_{C}^{2}}$ Applied voltage $\left(\mathrm{V}_{\mathrm{ap}}\right)=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}$ $=\sqrt{(5)^{2}+(12)^{2}}$ $\mathrm{~V}_{\text {ap }} =\sqrt{169}$ $\mathrm{~V}_{\mathrm{ap}} =13 \mathrm{~V}$ Hence, the applied voltage $\mathrm{V}_{\text {ap }}=13 \mathrm{~V}$.
Manipal UGET-2009
Alternating Current
155066
A L-R circuit consists of an inductance of $8 \mathrm{mH}$ and a resistance of $4 \Omega$. The time constant of the circuit is
1 $2 \mathrm{~ms}$
2 $12 \mathrm{~ms}$
3 $32 \mathrm{~ms}$
4 $500 \mathrm{~s}$
Explanation:
A Given that, Inductance $(\mathrm{L})=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=4 \Omega$ In L-R circuit, Time constant $(\mathrm{t})=\frac{\mathrm{L}}{\mathrm{R}}=\frac{8 \times 10^{-3}}{4}$ $=2 \times 10^{-3} \text { second }=2 \mathrm{~ms}$
