357477 In the given circuit, find charge on capacitor after \(1 s\) of opening the switch at \(t=\infty\).
1 \(20\,{e^{ - 10}}\mu C\)
2 \(25\,{e^{ - 10}}\mu C\)
3 \(30\,{e^{ - 10}}\mu C\)
4 \(35\,{e^{ - 10}}\mu C\)
Explanation:
At \(t \rightarrow \infty\), capacitor Works as on open circuit as shown in the figure. Therefore, current flow in the circuit, \({I_1} = \frac{9}{{(12 + 15){{10}^3}}} = \frac{1}{3} \times {10^{ - 3}}\;A\) \(\therefore {V_0} = {I_1} \times 15 \times {10^3}\) \( = \frac{1}{3} \times {10^{ - 3}} \times 15 \times {10^3} = 5\;V\) \(\therefore\) Charge on capacitor of \(5 \mu \mathrm{F}\),\(q_{0}=C V_{0}=5 \times 10^{-6} \times 5\) \(=25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C}\) When switch is open, then capacitor starts discharging across \((5+15=20 k \Omega)\) and instantaneous charge on capacitor is given by \(q=q_{0} e^{-t / R C}\) \(=25 e^{-\dfrac{1}{20 \times 10^{3} \times 5 \times 10^{-6}}}[\because t=1 \mathrm{~s}]\) \(=25 e^{-10} \mu C\)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357478
The capacitance of the system is \(C\); if the key is closed, the total energy loss is equal to:
1 \(\frac{{{Q^2}}}{{2C}}\)
2 \(\frac{{2{Q^2}}}{C}\)
3 \(\frac{{9{Q^2}}}{{8C}}\)
4 None of these
Explanation:
The charge distribution on the plates is Energy stored between the plates \( = \frac{{{{\left( {3{\text{Q}}/2} \right)}^2}}}{{2{\text{C}}}} = \frac{{9{{\text{Q}}^2}}}{{8{\text{C}}}}\) The entire energy which is stored between the plates is lost after closing the key-\(K\)
PHXII03:CURRENT ELECTRICITY
357479
In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
1 \(13\,\mu C\)
2 \(12\,\mu C\)
3 \(9\,\mu C\)
4 \({\rm{Zero}}\)
Explanation:
Sum of charges on plates of capacitor was zero initially and will be zero finally \(\therefore \) No charge flows to the ground.
PHXII03:CURRENT ELECTRICITY
357480
A galvanometer \((G)\) of \(2\,\Omega \) resistance is connected in the given circuit. The ratio of charges stored in \(C_{1}\) and \(C_{2}\) is
1 1
2 \(\dfrac{2}{3}\)
3 \(\dfrac{3}{2}\)
4 \(\dfrac{1}{2}\)
Explanation:
In steady state, capacitor behaves like an open circuit. So, \({R_{eq}} = 4 + 6 + 2 = 12\,\Omega \) Current, \(I = \frac{6}{{12}} \Rightarrow 0.5\;A\) Potential difference across \(C_{1}=\) Potential difference across \(4\,\Omega \)
357481
A capacitor of 8 \(F\) is connected as shown. Charge on the plates of the capacitor
1 \(32\,C\)
2 \(40\,C\)
3 \(0\,C\)
4 \(80\,C\)
Explanation:
At steady state, the capacitor gets fully charged and stops conducting. \(\therefore \) Current through the \(20\Omega \) resistance is zero \(\therefore \) Current in the circuit, \(I = \frac{5}{{4 + 1}} = 1A\) Voltage across the \(4\,\Omega \) resistance \( = V = I \times 4 = 4V\) = Voltage across the capacitor. Hence charge on capacitor, \(q = CV = 8 \times 4 = 32C\)
357477 In the given circuit, find charge on capacitor after \(1 s\) of opening the switch at \(t=\infty\).
1 \(20\,{e^{ - 10}}\mu C\)
2 \(25\,{e^{ - 10}}\mu C\)
3 \(30\,{e^{ - 10}}\mu C\)
4 \(35\,{e^{ - 10}}\mu C\)
Explanation:
At \(t \rightarrow \infty\), capacitor Works as on open circuit as shown in the figure. Therefore, current flow in the circuit, \({I_1} = \frac{9}{{(12 + 15){{10}^3}}} = \frac{1}{3} \times {10^{ - 3}}\;A\) \(\therefore {V_0} = {I_1} \times 15 \times {10^3}\) \( = \frac{1}{3} \times {10^{ - 3}} \times 15 \times {10^3} = 5\;V\) \(\therefore\) Charge on capacitor of \(5 \mu \mathrm{F}\),\(q_{0}=C V_{0}=5 \times 10^{-6} \times 5\) \(=25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C}\) When switch is open, then capacitor starts discharging across \((5+15=20 k \Omega)\) and instantaneous charge on capacitor is given by \(q=q_{0} e^{-t / R C}\) \(=25 e^{-\dfrac{1}{20 \times 10^{3} \times 5 \times 10^{-6}}}[\because t=1 \mathrm{~s}]\) \(=25 e^{-10} \mu C\)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357478
The capacitance of the system is \(C\); if the key is closed, the total energy loss is equal to:
1 \(\frac{{{Q^2}}}{{2C}}\)
2 \(\frac{{2{Q^2}}}{C}\)
3 \(\frac{{9{Q^2}}}{{8C}}\)
4 None of these
Explanation:
The charge distribution on the plates is Energy stored between the plates \( = \frac{{{{\left( {3{\text{Q}}/2} \right)}^2}}}{{2{\text{C}}}} = \frac{{9{{\text{Q}}^2}}}{{8{\text{C}}}}\) The entire energy which is stored between the plates is lost after closing the key-\(K\)
PHXII03:CURRENT ELECTRICITY
357479
In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
1 \(13\,\mu C\)
2 \(12\,\mu C\)
3 \(9\,\mu C\)
4 \({\rm{Zero}}\)
Explanation:
Sum of charges on plates of capacitor was zero initially and will be zero finally \(\therefore \) No charge flows to the ground.
PHXII03:CURRENT ELECTRICITY
357480
A galvanometer \((G)\) of \(2\,\Omega \) resistance is connected in the given circuit. The ratio of charges stored in \(C_{1}\) and \(C_{2}\) is
1 1
2 \(\dfrac{2}{3}\)
3 \(\dfrac{3}{2}\)
4 \(\dfrac{1}{2}\)
Explanation:
In steady state, capacitor behaves like an open circuit. So, \({R_{eq}} = 4 + 6 + 2 = 12\,\Omega \) Current, \(I = \frac{6}{{12}} \Rightarrow 0.5\;A\) Potential difference across \(C_{1}=\) Potential difference across \(4\,\Omega \)
357481
A capacitor of 8 \(F\) is connected as shown. Charge on the plates of the capacitor
1 \(32\,C\)
2 \(40\,C\)
3 \(0\,C\)
4 \(80\,C\)
Explanation:
At steady state, the capacitor gets fully charged and stops conducting. \(\therefore \) Current through the \(20\Omega \) resistance is zero \(\therefore \) Current in the circuit, \(I = \frac{5}{{4 + 1}} = 1A\) Voltage across the \(4\,\Omega \) resistance \( = V = I \times 4 = 4V\) = Voltage across the capacitor. Hence charge on capacitor, \(q = CV = 8 \times 4 = 32C\)
357477 In the given circuit, find charge on capacitor after \(1 s\) of opening the switch at \(t=\infty\).
1 \(20\,{e^{ - 10}}\mu C\)
2 \(25\,{e^{ - 10}}\mu C\)
3 \(30\,{e^{ - 10}}\mu C\)
4 \(35\,{e^{ - 10}}\mu C\)
Explanation:
At \(t \rightarrow \infty\), capacitor Works as on open circuit as shown in the figure. Therefore, current flow in the circuit, \({I_1} = \frac{9}{{(12 + 15){{10}^3}}} = \frac{1}{3} \times {10^{ - 3}}\;A\) \(\therefore {V_0} = {I_1} \times 15 \times {10^3}\) \( = \frac{1}{3} \times {10^{ - 3}} \times 15 \times {10^3} = 5\;V\) \(\therefore\) Charge on capacitor of \(5 \mu \mathrm{F}\),\(q_{0}=C V_{0}=5 \times 10^{-6} \times 5\) \(=25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C}\) When switch is open, then capacitor starts discharging across \((5+15=20 k \Omega)\) and instantaneous charge on capacitor is given by \(q=q_{0} e^{-t / R C}\) \(=25 e^{-\dfrac{1}{20 \times 10^{3} \times 5 \times 10^{-6}}}[\because t=1 \mathrm{~s}]\) \(=25 e^{-10} \mu C\)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357478
The capacitance of the system is \(C\); if the key is closed, the total energy loss is equal to:
1 \(\frac{{{Q^2}}}{{2C}}\)
2 \(\frac{{2{Q^2}}}{C}\)
3 \(\frac{{9{Q^2}}}{{8C}}\)
4 None of these
Explanation:
The charge distribution on the plates is Energy stored between the plates \( = \frac{{{{\left( {3{\text{Q}}/2} \right)}^2}}}{{2{\text{C}}}} = \frac{{9{{\text{Q}}^2}}}{{8{\text{C}}}}\) The entire energy which is stored between the plates is lost after closing the key-\(K\)
PHXII03:CURRENT ELECTRICITY
357479
In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
1 \(13\,\mu C\)
2 \(12\,\mu C\)
3 \(9\,\mu C\)
4 \({\rm{Zero}}\)
Explanation:
Sum of charges on plates of capacitor was zero initially and will be zero finally \(\therefore \) No charge flows to the ground.
PHXII03:CURRENT ELECTRICITY
357480
A galvanometer \((G)\) of \(2\,\Omega \) resistance is connected in the given circuit. The ratio of charges stored in \(C_{1}\) and \(C_{2}\) is
1 1
2 \(\dfrac{2}{3}\)
3 \(\dfrac{3}{2}\)
4 \(\dfrac{1}{2}\)
Explanation:
In steady state, capacitor behaves like an open circuit. So, \({R_{eq}} = 4 + 6 + 2 = 12\,\Omega \) Current, \(I = \frac{6}{{12}} \Rightarrow 0.5\;A\) Potential difference across \(C_{1}=\) Potential difference across \(4\,\Omega \)
357481
A capacitor of 8 \(F\) is connected as shown. Charge on the plates of the capacitor
1 \(32\,C\)
2 \(40\,C\)
3 \(0\,C\)
4 \(80\,C\)
Explanation:
At steady state, the capacitor gets fully charged and stops conducting. \(\therefore \) Current through the \(20\Omega \) resistance is zero \(\therefore \) Current in the circuit, \(I = \frac{5}{{4 + 1}} = 1A\) Voltage across the \(4\,\Omega \) resistance \( = V = I \times 4 = 4V\) = Voltage across the capacitor. Hence charge on capacitor, \(q = CV = 8 \times 4 = 32C\)
357477 In the given circuit, find charge on capacitor after \(1 s\) of opening the switch at \(t=\infty\).
1 \(20\,{e^{ - 10}}\mu C\)
2 \(25\,{e^{ - 10}}\mu C\)
3 \(30\,{e^{ - 10}}\mu C\)
4 \(35\,{e^{ - 10}}\mu C\)
Explanation:
At \(t \rightarrow \infty\), capacitor Works as on open circuit as shown in the figure. Therefore, current flow in the circuit, \({I_1} = \frac{9}{{(12 + 15){{10}^3}}} = \frac{1}{3} \times {10^{ - 3}}\;A\) \(\therefore {V_0} = {I_1} \times 15 \times {10^3}\) \( = \frac{1}{3} \times {10^{ - 3}} \times 15 \times {10^3} = 5\;V\) \(\therefore\) Charge on capacitor of \(5 \mu \mathrm{F}\),\(q_{0}=C V_{0}=5 \times 10^{-6} \times 5\) \(=25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C}\) When switch is open, then capacitor starts discharging across \((5+15=20 k \Omega)\) and instantaneous charge on capacitor is given by \(q=q_{0} e^{-t / R C}\) \(=25 e^{-\dfrac{1}{20 \times 10^{3} \times 5 \times 10^{-6}}}[\because t=1 \mathrm{~s}]\) \(=25 e^{-10} \mu C\)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357478
The capacitance of the system is \(C\); if the key is closed, the total energy loss is equal to:
1 \(\frac{{{Q^2}}}{{2C}}\)
2 \(\frac{{2{Q^2}}}{C}\)
3 \(\frac{{9{Q^2}}}{{8C}}\)
4 None of these
Explanation:
The charge distribution on the plates is Energy stored between the plates \( = \frac{{{{\left( {3{\text{Q}}/2} \right)}^2}}}{{2{\text{C}}}} = \frac{{9{{\text{Q}}^2}}}{{8{\text{C}}}}\) The entire energy which is stored between the plates is lost after closing the key-\(K\)
PHXII03:CURRENT ELECTRICITY
357479
In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
1 \(13\,\mu C\)
2 \(12\,\mu C\)
3 \(9\,\mu C\)
4 \({\rm{Zero}}\)
Explanation:
Sum of charges on plates of capacitor was zero initially and will be zero finally \(\therefore \) No charge flows to the ground.
PHXII03:CURRENT ELECTRICITY
357480
A galvanometer \((G)\) of \(2\,\Omega \) resistance is connected in the given circuit. The ratio of charges stored in \(C_{1}\) and \(C_{2}\) is
1 1
2 \(\dfrac{2}{3}\)
3 \(\dfrac{3}{2}\)
4 \(\dfrac{1}{2}\)
Explanation:
In steady state, capacitor behaves like an open circuit. So, \({R_{eq}} = 4 + 6 + 2 = 12\,\Omega \) Current, \(I = \frac{6}{{12}} \Rightarrow 0.5\;A\) Potential difference across \(C_{1}=\) Potential difference across \(4\,\Omega \)
357481
A capacitor of 8 \(F\) is connected as shown. Charge on the plates of the capacitor
1 \(32\,C\)
2 \(40\,C\)
3 \(0\,C\)
4 \(80\,C\)
Explanation:
At steady state, the capacitor gets fully charged and stops conducting. \(\therefore \) Current through the \(20\Omega \) resistance is zero \(\therefore \) Current in the circuit, \(I = \frac{5}{{4 + 1}} = 1A\) Voltage across the \(4\,\Omega \) resistance \( = V = I \times 4 = 4V\) = Voltage across the capacitor. Hence charge on capacitor, \(q = CV = 8 \times 4 = 32C\)
357477 In the given circuit, find charge on capacitor after \(1 s\) of opening the switch at \(t=\infty\).
1 \(20\,{e^{ - 10}}\mu C\)
2 \(25\,{e^{ - 10}}\mu C\)
3 \(30\,{e^{ - 10}}\mu C\)
4 \(35\,{e^{ - 10}}\mu C\)
Explanation:
At \(t \rightarrow \infty\), capacitor Works as on open circuit as shown in the figure. Therefore, current flow in the circuit, \({I_1} = \frac{9}{{(12 + 15){{10}^3}}} = \frac{1}{3} \times {10^{ - 3}}\;A\) \(\therefore {V_0} = {I_1} \times 15 \times {10^3}\) \( = \frac{1}{3} \times {10^{ - 3}} \times 15 \times {10^3} = 5\;V\) \(\therefore\) Charge on capacitor of \(5 \mu \mathrm{F}\),\(q_{0}=C V_{0}=5 \times 10^{-6} \times 5\) \(=25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C}\) When switch is open, then capacitor starts discharging across \((5+15=20 k \Omega)\) and instantaneous charge on capacitor is given by \(q=q_{0} e^{-t / R C}\) \(=25 e^{-\dfrac{1}{20 \times 10^{3} \times 5 \times 10^{-6}}}[\because t=1 \mathrm{~s}]\) \(=25 e^{-10} \mu C\)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357478
The capacitance of the system is \(C\); if the key is closed, the total energy loss is equal to:
1 \(\frac{{{Q^2}}}{{2C}}\)
2 \(\frac{{2{Q^2}}}{C}\)
3 \(\frac{{9{Q^2}}}{{8C}}\)
4 None of these
Explanation:
The charge distribution on the plates is Energy stored between the plates \( = \frac{{{{\left( {3{\text{Q}}/2} \right)}^2}}}{{2{\text{C}}}} = \frac{{9{{\text{Q}}^2}}}{{8{\text{C}}}}\) The entire energy which is stored between the plates is lost after closing the key-\(K\)
PHXII03:CURRENT ELECTRICITY
357479
In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
1 \(13\,\mu C\)
2 \(12\,\mu C\)
3 \(9\,\mu C\)
4 \({\rm{Zero}}\)
Explanation:
Sum of charges on plates of capacitor was zero initially and will be zero finally \(\therefore \) No charge flows to the ground.
PHXII03:CURRENT ELECTRICITY
357480
A galvanometer \((G)\) of \(2\,\Omega \) resistance is connected in the given circuit. The ratio of charges stored in \(C_{1}\) and \(C_{2}\) is
1 1
2 \(\dfrac{2}{3}\)
3 \(\dfrac{3}{2}\)
4 \(\dfrac{1}{2}\)
Explanation:
In steady state, capacitor behaves like an open circuit. So, \({R_{eq}} = 4 + 6 + 2 = 12\,\Omega \) Current, \(I = \frac{6}{{12}} \Rightarrow 0.5\;A\) Potential difference across \(C_{1}=\) Potential difference across \(4\,\Omega \)
357481
A capacitor of 8 \(F\) is connected as shown. Charge on the plates of the capacitor
1 \(32\,C\)
2 \(40\,C\)
3 \(0\,C\)
4 \(80\,C\)
Explanation:
At steady state, the capacitor gets fully charged and stops conducting. \(\therefore \) Current through the \(20\Omega \) resistance is zero \(\therefore \) Current in the circuit, \(I = \frac{5}{{4 + 1}} = 1A\) Voltage across the \(4\,\Omega \) resistance \( = V = I \times 4 = 4V\) = Voltage across the capacitor. Hence charge on capacitor, \(q = CV = 8 \times 4 = 32C\)
