153090
In the network shown below, the charge accumulated in the capacitor in steady state will be :
1 $7.2 \mu \mathrm{C}$
2 $4.8 \mu \mathrm{C}$
3 $10.3 \mu \mathrm{C}$
4 $12 \mu \mathrm{C}$
Explanation:
A We know that, in steady state, no current will flow in capacitor. $\because$ Potential difference on $6 \Omega$ resistance - $\mathrm{V}=6 \times \frac{3}{10}=1.8 \text { Volt }$ Therefore, capacitor will have same potential So, charge $(q)=C V$ $=4 \mu \mathrm{F} \times 1.8 \text { volt }=7.2 \mu \mathrm{C} \text {. }$
JEE Main-13.04.2023
Current Electricity
153091
What are the values of ' $E$ ' in the following circuit, if a current of 2 A flows in the clockwise as well in anticlockwise direction?
1 $3 \mathrm{~V}, 28 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $3 \mathrm{~V}, 30 \mathrm{~V}$
4 $3 \mathrm{~V}, 2.8 \mathrm{~V}$
Explanation:
B From figure (I), When $2 \mathrm{~A}$ current is following clockwise apply Kirchoff's low, When, current flow in clockwise direction. From figure (II), $E+(6 \times 2)-20+3 \times 2=0$ $E+12-20+6=0$ $E=2 V$ When, current flow in anticlockwise direction.
AP EAMCET-05.10.2021
Current Electricity
153092
An emf of $15 \mathrm{~V}$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of the currents at time, $t=\infty$ and $t=1$ s is
1 $\frac{\mathrm{e}}{\mathrm{e}^{2}-1}$
2 $\frac{\mathrm{e}^{2}}{\mathrm{e}-1}$
3 $\frac{\mathrm{e}}{1-\mathrm{e}^{2}}$
4 $\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
Explanation:
D Given, $\mathrm{V}_{0}=12 \mathrm{~V}, \mathrm{R}=10 \Omega, \mathrm{L}=5 \mathrm{H}$ The current in RL circuit, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{5}{10}=\frac{1}{2} \mathrm{sec}$ We know that, $\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}\right)$ For $\mathrm{t}=\infty \mathrm{sec}$, For $\mathrm{t}=1 \mathrm{I}=\mathrm{I}_{0}$ $\mathrm{I}_{1}=\mathrm{I}_{0}$ $1 \mathrm{sec}$ $\mathrm{I}_{2}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-2}\right)$ Ratio of currents at time $t=\infty$ and $t=1 \mathrm{sec}$ $\frac{I_{1}}{I_{2}}=\frac{I_{0}}{I_{0}\left(1-\mathrm{e}^{-2}\right)}$ $\frac{I_{1}}{I_{2}}=\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
AP EAMCET (21.04.2019) Shift-II
Current Electricity
153093 Initially the switch is closed till the steady state is reached. Now find charge on capacitor after 1 sec of opening the switch.
1 $20 \mathrm{e}^{-10} \mu \mathrm{C}$
2 $25 \mathrm{e}^{-10} \mu \mathrm{C}$
3 $30 \mathrm{e}^{-10} \mu \mathrm{C}$
4 $35 \mathrm{e}^{-10} \mu \mathrm{C}$
Explanation:
B Capacitor is in steady state, So current through the circuit $\mathrm{i}=\frac{9}{(12+15) \times 10^{3}}=\frac{10^{-3}}{3}$ Potential at $15 \mathrm{k} \Omega$ resistance, $\mathrm{V}=\mathrm{iR}$ $\mathrm{V}=\frac{10^{-3}}{3} \times 15 \times 10^{3}$ $\mathrm{V}=5 \text { volt }$ Charge, $\mathrm{q}_{\mathrm{o}}=\mathrm{CV}$ $\mathrm{q}_{\mathrm{o}}=5 \times 5$ $\mathrm{q}_{\mathrm{o}}=25 \mu \mathrm{C}$ When, switch is opened Resistance of above circuit $(\mathrm{R})=15+5=20 \mathrm{k} \Omega$ $\mathrm{R}=20 \times 10^{-3} \Omega$ During discharging charge on capacitor $\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} / R C}$ $\mathrm{q}=25 \times \mathrm{e}^{\frac{-1}{20 \times 10^{3} \times 5 \times 10^{-6}}}$ $\mathrm{q}=25 \mathrm{e}^{-10} \mu \mathrm{C}$
AIIMS-25.05.2019(E) Shift-2
Current Electricity
153095
In steady state, charge on $3 \mu \mathrm{F}$ capacitor is
1 $36 \mu \mathrm{C}$
2 $27 \mu \mathrm{C}$
3 $18 \mu \mathrm{C}$
4 $54 \mu \mathrm{C}$
Explanation:
B The circuit diagram as following current in the circuit $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $i =\frac{18}{6+6}$ $=1.5 \mathrm{~A}$ Voltage drop across $\mathrm{R}_{2}$, $\mathrm{V}_{2} =\mathrm{R}_{2} \mathrm{i}$ $=6 \times 1.5$ $=9 \mathrm{~V}$ Charge on capacitor $\left(\mathrm{C}_{2}=3 \mu \mathrm{F}\right)$, $\mathrm{C}_{2} \mathrm{~V}_{2} =3 \times 9$ $=27 \mu \mathrm{C}$
153090
In the network shown below, the charge accumulated in the capacitor in steady state will be :
1 $7.2 \mu \mathrm{C}$
2 $4.8 \mu \mathrm{C}$
3 $10.3 \mu \mathrm{C}$
4 $12 \mu \mathrm{C}$
Explanation:
A We know that, in steady state, no current will flow in capacitor. $\because$ Potential difference on $6 \Omega$ resistance - $\mathrm{V}=6 \times \frac{3}{10}=1.8 \text { Volt }$ Therefore, capacitor will have same potential So, charge $(q)=C V$ $=4 \mu \mathrm{F} \times 1.8 \text { volt }=7.2 \mu \mathrm{C} \text {. }$
JEE Main-13.04.2023
Current Electricity
153091
What are the values of ' $E$ ' in the following circuit, if a current of 2 A flows in the clockwise as well in anticlockwise direction?
1 $3 \mathrm{~V}, 28 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $3 \mathrm{~V}, 30 \mathrm{~V}$
4 $3 \mathrm{~V}, 2.8 \mathrm{~V}$
Explanation:
B From figure (I), When $2 \mathrm{~A}$ current is following clockwise apply Kirchoff's low, When, current flow in clockwise direction. From figure (II), $E+(6 \times 2)-20+3 \times 2=0$ $E+12-20+6=0$ $E=2 V$ When, current flow in anticlockwise direction.
AP EAMCET-05.10.2021
Current Electricity
153092
An emf of $15 \mathrm{~V}$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of the currents at time, $t=\infty$ and $t=1$ s is
1 $\frac{\mathrm{e}}{\mathrm{e}^{2}-1}$
2 $\frac{\mathrm{e}^{2}}{\mathrm{e}-1}$
3 $\frac{\mathrm{e}}{1-\mathrm{e}^{2}}$
4 $\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
Explanation:
D Given, $\mathrm{V}_{0}=12 \mathrm{~V}, \mathrm{R}=10 \Omega, \mathrm{L}=5 \mathrm{H}$ The current in RL circuit, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{5}{10}=\frac{1}{2} \mathrm{sec}$ We know that, $\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}\right)$ For $\mathrm{t}=\infty \mathrm{sec}$, For $\mathrm{t}=1 \mathrm{I}=\mathrm{I}_{0}$ $\mathrm{I}_{1}=\mathrm{I}_{0}$ $1 \mathrm{sec}$ $\mathrm{I}_{2}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-2}\right)$ Ratio of currents at time $t=\infty$ and $t=1 \mathrm{sec}$ $\frac{I_{1}}{I_{2}}=\frac{I_{0}}{I_{0}\left(1-\mathrm{e}^{-2}\right)}$ $\frac{I_{1}}{I_{2}}=\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
AP EAMCET (21.04.2019) Shift-II
Current Electricity
153093 Initially the switch is closed till the steady state is reached. Now find charge on capacitor after 1 sec of opening the switch.
1 $20 \mathrm{e}^{-10} \mu \mathrm{C}$
2 $25 \mathrm{e}^{-10} \mu \mathrm{C}$
3 $30 \mathrm{e}^{-10} \mu \mathrm{C}$
4 $35 \mathrm{e}^{-10} \mu \mathrm{C}$
Explanation:
B Capacitor is in steady state, So current through the circuit $\mathrm{i}=\frac{9}{(12+15) \times 10^{3}}=\frac{10^{-3}}{3}$ Potential at $15 \mathrm{k} \Omega$ resistance, $\mathrm{V}=\mathrm{iR}$ $\mathrm{V}=\frac{10^{-3}}{3} \times 15 \times 10^{3}$ $\mathrm{V}=5 \text { volt }$ Charge, $\mathrm{q}_{\mathrm{o}}=\mathrm{CV}$ $\mathrm{q}_{\mathrm{o}}=5 \times 5$ $\mathrm{q}_{\mathrm{o}}=25 \mu \mathrm{C}$ When, switch is opened Resistance of above circuit $(\mathrm{R})=15+5=20 \mathrm{k} \Omega$ $\mathrm{R}=20 \times 10^{-3} \Omega$ During discharging charge on capacitor $\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} / R C}$ $\mathrm{q}=25 \times \mathrm{e}^{\frac{-1}{20 \times 10^{3} \times 5 \times 10^{-6}}}$ $\mathrm{q}=25 \mathrm{e}^{-10} \mu \mathrm{C}$
AIIMS-25.05.2019(E) Shift-2
Current Electricity
153095
In steady state, charge on $3 \mu \mathrm{F}$ capacitor is
1 $36 \mu \mathrm{C}$
2 $27 \mu \mathrm{C}$
3 $18 \mu \mathrm{C}$
4 $54 \mu \mathrm{C}$
Explanation:
B The circuit diagram as following current in the circuit $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $i =\frac{18}{6+6}$ $=1.5 \mathrm{~A}$ Voltage drop across $\mathrm{R}_{2}$, $\mathrm{V}_{2} =\mathrm{R}_{2} \mathrm{i}$ $=6 \times 1.5$ $=9 \mathrm{~V}$ Charge on capacitor $\left(\mathrm{C}_{2}=3 \mu \mathrm{F}\right)$, $\mathrm{C}_{2} \mathrm{~V}_{2} =3 \times 9$ $=27 \mu \mathrm{C}$
153090
In the network shown below, the charge accumulated in the capacitor in steady state will be :
1 $7.2 \mu \mathrm{C}$
2 $4.8 \mu \mathrm{C}$
3 $10.3 \mu \mathrm{C}$
4 $12 \mu \mathrm{C}$
Explanation:
A We know that, in steady state, no current will flow in capacitor. $\because$ Potential difference on $6 \Omega$ resistance - $\mathrm{V}=6 \times \frac{3}{10}=1.8 \text { Volt }$ Therefore, capacitor will have same potential So, charge $(q)=C V$ $=4 \mu \mathrm{F} \times 1.8 \text { volt }=7.2 \mu \mathrm{C} \text {. }$
JEE Main-13.04.2023
Current Electricity
153091
What are the values of ' $E$ ' in the following circuit, if a current of 2 A flows in the clockwise as well in anticlockwise direction?
1 $3 \mathrm{~V}, 28 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $3 \mathrm{~V}, 30 \mathrm{~V}$
4 $3 \mathrm{~V}, 2.8 \mathrm{~V}$
Explanation:
B From figure (I), When $2 \mathrm{~A}$ current is following clockwise apply Kirchoff's low, When, current flow in clockwise direction. From figure (II), $E+(6 \times 2)-20+3 \times 2=0$ $E+12-20+6=0$ $E=2 V$ When, current flow in anticlockwise direction.
AP EAMCET-05.10.2021
Current Electricity
153092
An emf of $15 \mathrm{~V}$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of the currents at time, $t=\infty$ and $t=1$ s is
1 $\frac{\mathrm{e}}{\mathrm{e}^{2}-1}$
2 $\frac{\mathrm{e}^{2}}{\mathrm{e}-1}$
3 $\frac{\mathrm{e}}{1-\mathrm{e}^{2}}$
4 $\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
Explanation:
D Given, $\mathrm{V}_{0}=12 \mathrm{~V}, \mathrm{R}=10 \Omega, \mathrm{L}=5 \mathrm{H}$ The current in RL circuit, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{5}{10}=\frac{1}{2} \mathrm{sec}$ We know that, $\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}\right)$ For $\mathrm{t}=\infty \mathrm{sec}$, For $\mathrm{t}=1 \mathrm{I}=\mathrm{I}_{0}$ $\mathrm{I}_{1}=\mathrm{I}_{0}$ $1 \mathrm{sec}$ $\mathrm{I}_{2}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-2}\right)$ Ratio of currents at time $t=\infty$ and $t=1 \mathrm{sec}$ $\frac{I_{1}}{I_{2}}=\frac{I_{0}}{I_{0}\left(1-\mathrm{e}^{-2}\right)}$ $\frac{I_{1}}{I_{2}}=\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
AP EAMCET (21.04.2019) Shift-II
Current Electricity
153093 Initially the switch is closed till the steady state is reached. Now find charge on capacitor after 1 sec of opening the switch.
1 $20 \mathrm{e}^{-10} \mu \mathrm{C}$
2 $25 \mathrm{e}^{-10} \mu \mathrm{C}$
3 $30 \mathrm{e}^{-10} \mu \mathrm{C}$
4 $35 \mathrm{e}^{-10} \mu \mathrm{C}$
Explanation:
B Capacitor is in steady state, So current through the circuit $\mathrm{i}=\frac{9}{(12+15) \times 10^{3}}=\frac{10^{-3}}{3}$ Potential at $15 \mathrm{k} \Omega$ resistance, $\mathrm{V}=\mathrm{iR}$ $\mathrm{V}=\frac{10^{-3}}{3} \times 15 \times 10^{3}$ $\mathrm{V}=5 \text { volt }$ Charge, $\mathrm{q}_{\mathrm{o}}=\mathrm{CV}$ $\mathrm{q}_{\mathrm{o}}=5 \times 5$ $\mathrm{q}_{\mathrm{o}}=25 \mu \mathrm{C}$ When, switch is opened Resistance of above circuit $(\mathrm{R})=15+5=20 \mathrm{k} \Omega$ $\mathrm{R}=20 \times 10^{-3} \Omega$ During discharging charge on capacitor $\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} / R C}$ $\mathrm{q}=25 \times \mathrm{e}^{\frac{-1}{20 \times 10^{3} \times 5 \times 10^{-6}}}$ $\mathrm{q}=25 \mathrm{e}^{-10} \mu \mathrm{C}$
AIIMS-25.05.2019(E) Shift-2
Current Electricity
153095
In steady state, charge on $3 \mu \mathrm{F}$ capacitor is
1 $36 \mu \mathrm{C}$
2 $27 \mu \mathrm{C}$
3 $18 \mu \mathrm{C}$
4 $54 \mu \mathrm{C}$
Explanation:
B The circuit diagram as following current in the circuit $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $i =\frac{18}{6+6}$ $=1.5 \mathrm{~A}$ Voltage drop across $\mathrm{R}_{2}$, $\mathrm{V}_{2} =\mathrm{R}_{2} \mathrm{i}$ $=6 \times 1.5$ $=9 \mathrm{~V}$ Charge on capacitor $\left(\mathrm{C}_{2}=3 \mu \mathrm{F}\right)$, $\mathrm{C}_{2} \mathrm{~V}_{2} =3 \times 9$ $=27 \mu \mathrm{C}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
153090
In the network shown below, the charge accumulated in the capacitor in steady state will be :
1 $7.2 \mu \mathrm{C}$
2 $4.8 \mu \mathrm{C}$
3 $10.3 \mu \mathrm{C}$
4 $12 \mu \mathrm{C}$
Explanation:
A We know that, in steady state, no current will flow in capacitor. $\because$ Potential difference on $6 \Omega$ resistance - $\mathrm{V}=6 \times \frac{3}{10}=1.8 \text { Volt }$ Therefore, capacitor will have same potential So, charge $(q)=C V$ $=4 \mu \mathrm{F} \times 1.8 \text { volt }=7.2 \mu \mathrm{C} \text {. }$
JEE Main-13.04.2023
Current Electricity
153091
What are the values of ' $E$ ' in the following circuit, if a current of 2 A flows in the clockwise as well in anticlockwise direction?
1 $3 \mathrm{~V}, 28 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $3 \mathrm{~V}, 30 \mathrm{~V}$
4 $3 \mathrm{~V}, 2.8 \mathrm{~V}$
Explanation:
B From figure (I), When $2 \mathrm{~A}$ current is following clockwise apply Kirchoff's low, When, current flow in clockwise direction. From figure (II), $E+(6 \times 2)-20+3 \times 2=0$ $E+12-20+6=0$ $E=2 V$ When, current flow in anticlockwise direction.
AP EAMCET-05.10.2021
Current Electricity
153092
An emf of $15 \mathrm{~V}$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of the currents at time, $t=\infty$ and $t=1$ s is
1 $\frac{\mathrm{e}}{\mathrm{e}^{2}-1}$
2 $\frac{\mathrm{e}^{2}}{\mathrm{e}-1}$
3 $\frac{\mathrm{e}}{1-\mathrm{e}^{2}}$
4 $\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
Explanation:
D Given, $\mathrm{V}_{0}=12 \mathrm{~V}, \mathrm{R}=10 \Omega, \mathrm{L}=5 \mathrm{H}$ The current in RL circuit, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{5}{10}=\frac{1}{2} \mathrm{sec}$ We know that, $\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}\right)$ For $\mathrm{t}=\infty \mathrm{sec}$, For $\mathrm{t}=1 \mathrm{I}=\mathrm{I}_{0}$ $\mathrm{I}_{1}=\mathrm{I}_{0}$ $1 \mathrm{sec}$ $\mathrm{I}_{2}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-2}\right)$ Ratio of currents at time $t=\infty$ and $t=1 \mathrm{sec}$ $\frac{I_{1}}{I_{2}}=\frac{I_{0}}{I_{0}\left(1-\mathrm{e}^{-2}\right)}$ $\frac{I_{1}}{I_{2}}=\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
AP EAMCET (21.04.2019) Shift-II
Current Electricity
153093 Initially the switch is closed till the steady state is reached. Now find charge on capacitor after 1 sec of opening the switch.
1 $20 \mathrm{e}^{-10} \mu \mathrm{C}$
2 $25 \mathrm{e}^{-10} \mu \mathrm{C}$
3 $30 \mathrm{e}^{-10} \mu \mathrm{C}$
4 $35 \mathrm{e}^{-10} \mu \mathrm{C}$
Explanation:
B Capacitor is in steady state, So current through the circuit $\mathrm{i}=\frac{9}{(12+15) \times 10^{3}}=\frac{10^{-3}}{3}$ Potential at $15 \mathrm{k} \Omega$ resistance, $\mathrm{V}=\mathrm{iR}$ $\mathrm{V}=\frac{10^{-3}}{3} \times 15 \times 10^{3}$ $\mathrm{V}=5 \text { volt }$ Charge, $\mathrm{q}_{\mathrm{o}}=\mathrm{CV}$ $\mathrm{q}_{\mathrm{o}}=5 \times 5$ $\mathrm{q}_{\mathrm{o}}=25 \mu \mathrm{C}$ When, switch is opened Resistance of above circuit $(\mathrm{R})=15+5=20 \mathrm{k} \Omega$ $\mathrm{R}=20 \times 10^{-3} \Omega$ During discharging charge on capacitor $\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} / R C}$ $\mathrm{q}=25 \times \mathrm{e}^{\frac{-1}{20 \times 10^{3} \times 5 \times 10^{-6}}}$ $\mathrm{q}=25 \mathrm{e}^{-10} \mu \mathrm{C}$
AIIMS-25.05.2019(E) Shift-2
Current Electricity
153095
In steady state, charge on $3 \mu \mathrm{F}$ capacitor is
1 $36 \mu \mathrm{C}$
2 $27 \mu \mathrm{C}$
3 $18 \mu \mathrm{C}$
4 $54 \mu \mathrm{C}$
Explanation:
B The circuit diagram as following current in the circuit $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $i =\frac{18}{6+6}$ $=1.5 \mathrm{~A}$ Voltage drop across $\mathrm{R}_{2}$, $\mathrm{V}_{2} =\mathrm{R}_{2} \mathrm{i}$ $=6 \times 1.5$ $=9 \mathrm{~V}$ Charge on capacitor $\left(\mathrm{C}_{2}=3 \mu \mathrm{F}\right)$, $\mathrm{C}_{2} \mathrm{~V}_{2} =3 \times 9$ $=27 \mu \mathrm{C}$
153090
In the network shown below, the charge accumulated in the capacitor in steady state will be :
1 $7.2 \mu \mathrm{C}$
2 $4.8 \mu \mathrm{C}$
3 $10.3 \mu \mathrm{C}$
4 $12 \mu \mathrm{C}$
Explanation:
A We know that, in steady state, no current will flow in capacitor. $\because$ Potential difference on $6 \Omega$ resistance - $\mathrm{V}=6 \times \frac{3}{10}=1.8 \text { Volt }$ Therefore, capacitor will have same potential So, charge $(q)=C V$ $=4 \mu \mathrm{F} \times 1.8 \text { volt }=7.2 \mu \mathrm{C} \text {. }$
JEE Main-13.04.2023
Current Electricity
153091
What are the values of ' $E$ ' in the following circuit, if a current of 2 A flows in the clockwise as well in anticlockwise direction?
1 $3 \mathrm{~V}, 28 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $3 \mathrm{~V}, 30 \mathrm{~V}$
4 $3 \mathrm{~V}, 2.8 \mathrm{~V}$
Explanation:
B From figure (I), When $2 \mathrm{~A}$ current is following clockwise apply Kirchoff's low, When, current flow in clockwise direction. From figure (II), $E+(6 \times 2)-20+3 \times 2=0$ $E+12-20+6=0$ $E=2 V$ When, current flow in anticlockwise direction.
AP EAMCET-05.10.2021
Current Electricity
153092
An emf of $15 \mathrm{~V}$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of the currents at time, $t=\infty$ and $t=1$ s is
1 $\frac{\mathrm{e}}{\mathrm{e}^{2}-1}$
2 $\frac{\mathrm{e}^{2}}{\mathrm{e}-1}$
3 $\frac{\mathrm{e}}{1-\mathrm{e}^{2}}$
4 $\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
Explanation:
D Given, $\mathrm{V}_{0}=12 \mathrm{~V}, \mathrm{R}=10 \Omega, \mathrm{L}=5 \mathrm{H}$ The current in RL circuit, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{5}{10}=\frac{1}{2} \mathrm{sec}$ We know that, $\mathrm{i}=\mathrm{i}_{0}\left(1-\mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}\right)$ For $\mathrm{t}=\infty \mathrm{sec}$, For $\mathrm{t}=1 \mathrm{I}=\mathrm{I}_{0}$ $\mathrm{I}_{1}=\mathrm{I}_{0}$ $1 \mathrm{sec}$ $\mathrm{I}_{2}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-2}\right)$ Ratio of currents at time $t=\infty$ and $t=1 \mathrm{sec}$ $\frac{I_{1}}{I_{2}}=\frac{I_{0}}{I_{0}\left(1-\mathrm{e}^{-2}\right)}$ $\frac{I_{1}}{I_{2}}=\frac{\mathrm{e}^{2}}{\mathrm{e}^{2}-1}$
AP EAMCET (21.04.2019) Shift-II
Current Electricity
153093 Initially the switch is closed till the steady state is reached. Now find charge on capacitor after 1 sec of opening the switch.
1 $20 \mathrm{e}^{-10} \mu \mathrm{C}$
2 $25 \mathrm{e}^{-10} \mu \mathrm{C}$
3 $30 \mathrm{e}^{-10} \mu \mathrm{C}$
4 $35 \mathrm{e}^{-10} \mu \mathrm{C}$
Explanation:
B Capacitor is in steady state, So current through the circuit $\mathrm{i}=\frac{9}{(12+15) \times 10^{3}}=\frac{10^{-3}}{3}$ Potential at $15 \mathrm{k} \Omega$ resistance, $\mathrm{V}=\mathrm{iR}$ $\mathrm{V}=\frac{10^{-3}}{3} \times 15 \times 10^{3}$ $\mathrm{V}=5 \text { volt }$ Charge, $\mathrm{q}_{\mathrm{o}}=\mathrm{CV}$ $\mathrm{q}_{\mathrm{o}}=5 \times 5$ $\mathrm{q}_{\mathrm{o}}=25 \mu \mathrm{C}$ When, switch is opened Resistance of above circuit $(\mathrm{R})=15+5=20 \mathrm{k} \Omega$ $\mathrm{R}=20 \times 10^{-3} \Omega$ During discharging charge on capacitor $\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} / R C}$ $\mathrm{q}=25 \times \mathrm{e}^{\frac{-1}{20 \times 10^{3} \times 5 \times 10^{-6}}}$ $\mathrm{q}=25 \mathrm{e}^{-10} \mu \mathrm{C}$
AIIMS-25.05.2019(E) Shift-2
Current Electricity
153095
In steady state, charge on $3 \mu \mathrm{F}$ capacitor is
1 $36 \mu \mathrm{C}$
2 $27 \mu \mathrm{C}$
3 $18 \mu \mathrm{C}$
4 $54 \mu \mathrm{C}$
Explanation:
B The circuit diagram as following current in the circuit $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $i =\frac{18}{6+6}$ $=1.5 \mathrm{~A}$ Voltage drop across $\mathrm{R}_{2}$, $\mathrm{V}_{2} =\mathrm{R}_{2} \mathrm{i}$ $=6 \times 1.5$ $=9 \mathrm{~V}$ Charge on capacitor $\left(\mathrm{C}_{2}=3 \mu \mathrm{F}\right)$, $\mathrm{C}_{2} \mathrm{~V}_{2} =3 \times 9$ $=27 \mu \mathrm{C}$
