359290
The effective capacitance between the point \(P\) and \(Q\) in the given figure is
1 \(4\;\mu F\)
2 \(16\;\mu F\)
3 \(26\;\mu F\)
4 \(10\;\mu F\)
Explanation:
The points having equal potentials are shown in the figure The circuit can be redrawn as The given circuit is a wheatstone bridge. \({C_{PQ}} = 4\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359291
When a potential difference of \({10^3}V\) is applied between \(A\) and \(B\), a charge of 0.75 \(mC\) is stored in the system of capacitors as shown. The value of \(C\) is (in \(\mu F\))
1 \(\frac{1}{2}\)
2 \(2\)
3 \(2.5\)
4 \(3\)
Explanation:
The equivalent circuit of the given network is as shown in the figures. Hence, the equivalent capacitance between \(A\) and \(B\) \({C_{eq}}\; = \;\frac{{1\left( {C\; + \;1} \right)}}{{1\; + \;C\; + \;1}}\; = \;\frac{{C\; + \;1}}{{C\; + \;2}}\) The charge stored on the system of capacitors is \(Q = 0.75mC = 0.75 \times {10^{ - 3}}C\) As \(Q = {C_{eq}}V\) \(0.75\; \times \;{10^{ - 3}}\; = \;\frac{{\left( {C\; + \;1} \right)}}{{\left( {C\; + \;2} \right)}}\; \times \;{10^{ - 6}} \times \;{10^3}\) On solving, we get \(C = 2\)
KCET - 2013
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359292
Three capacitors each of \(4\,\mu F\) are to be connected in such a way that the effective capacitance is \(6\,\mu F\). This can be done by connecting them:
1 All in series
2 Two in parallel and one in series
3 Two in series and one in parallel
4 All in parallel
Explanation:
Conceptual Question
JEE - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359293
For the arrangement of capacitors as shown in the circuit, the effective capacitance between the points \(A\) and \(B\) is (capacitance of each capacitor is \(4\mu F\))
1 \(4\mu F\)
2 \(2\mu F\)
3 \(1\mu F\)
4 \(8\mu F\)
Explanation:
The given circuit can be redrawn as shown in figure (ii) As \(\frac{{{C_{AC}}}}{{{C_{AD}}}} = \frac{{{C_{CB}}}}{{{C_{DB}}}}\) The given circuit is a balanced Wheatstone bridge and the capacitance in arm CD is ineffective. Thus, it reduces to the equivalent circuit as shown in figure (iii). As \(4\mu C\) and \(4\mu C\) are in series in the upper branch, their equivalent capacitance \({C_1}\) is \({C_1}\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{4\mu F\; + \;4\mu F}}\; = \;2\mu F\) Similarly, \({4\mu F}\) and \({4\mu F}\) are in series in the lower branch, so their equivalent capacitance \({C_2}\) is \({C_2}\;\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{C\; + \;4\mu F}}\; = \;2\mu F\) And the corresponding equivalent circuit is shown in figure (iv). Since \({C_1}\,{\rm{and}}\,{C_2}\) are in parallel, the effective capacitance between the points A and B is \({C_{AB}} = {C_1} + {C_2} = 2\mu F + 2\mu F = 4\mu F\)
359290
The effective capacitance between the point \(P\) and \(Q\) in the given figure is
1 \(4\;\mu F\)
2 \(16\;\mu F\)
3 \(26\;\mu F\)
4 \(10\;\mu F\)
Explanation:
The points having equal potentials are shown in the figure The circuit can be redrawn as The given circuit is a wheatstone bridge. \({C_{PQ}} = 4\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359291
When a potential difference of \({10^3}V\) is applied between \(A\) and \(B\), a charge of 0.75 \(mC\) is stored in the system of capacitors as shown. The value of \(C\) is (in \(\mu F\))
1 \(\frac{1}{2}\)
2 \(2\)
3 \(2.5\)
4 \(3\)
Explanation:
The equivalent circuit of the given network is as shown in the figures. Hence, the equivalent capacitance between \(A\) and \(B\) \({C_{eq}}\; = \;\frac{{1\left( {C\; + \;1} \right)}}{{1\; + \;C\; + \;1}}\; = \;\frac{{C\; + \;1}}{{C\; + \;2}}\) The charge stored on the system of capacitors is \(Q = 0.75mC = 0.75 \times {10^{ - 3}}C\) As \(Q = {C_{eq}}V\) \(0.75\; \times \;{10^{ - 3}}\; = \;\frac{{\left( {C\; + \;1} \right)}}{{\left( {C\; + \;2} \right)}}\; \times \;{10^{ - 6}} \times \;{10^3}\) On solving, we get \(C = 2\)
KCET - 2013
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359292
Three capacitors each of \(4\,\mu F\) are to be connected in such a way that the effective capacitance is \(6\,\mu F\). This can be done by connecting them:
1 All in series
2 Two in parallel and one in series
3 Two in series and one in parallel
4 All in parallel
Explanation:
Conceptual Question
JEE - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359293
For the arrangement of capacitors as shown in the circuit, the effective capacitance between the points \(A\) and \(B\) is (capacitance of each capacitor is \(4\mu F\))
1 \(4\mu F\)
2 \(2\mu F\)
3 \(1\mu F\)
4 \(8\mu F\)
Explanation:
The given circuit can be redrawn as shown in figure (ii) As \(\frac{{{C_{AC}}}}{{{C_{AD}}}} = \frac{{{C_{CB}}}}{{{C_{DB}}}}\) The given circuit is a balanced Wheatstone bridge and the capacitance in arm CD is ineffective. Thus, it reduces to the equivalent circuit as shown in figure (iii). As \(4\mu C\) and \(4\mu C\) are in series in the upper branch, their equivalent capacitance \({C_1}\) is \({C_1}\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{4\mu F\; + \;4\mu F}}\; = \;2\mu F\) Similarly, \({4\mu F}\) and \({4\mu F}\) are in series in the lower branch, so their equivalent capacitance \({C_2}\) is \({C_2}\;\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{C\; + \;4\mu F}}\; = \;2\mu F\) And the corresponding equivalent circuit is shown in figure (iv). Since \({C_1}\,{\rm{and}}\,{C_2}\) are in parallel, the effective capacitance between the points A and B is \({C_{AB}} = {C_1} + {C_2} = 2\mu F + 2\mu F = 4\mu F\)
359290
The effective capacitance between the point \(P\) and \(Q\) in the given figure is
1 \(4\;\mu F\)
2 \(16\;\mu F\)
3 \(26\;\mu F\)
4 \(10\;\mu F\)
Explanation:
The points having equal potentials are shown in the figure The circuit can be redrawn as The given circuit is a wheatstone bridge. \({C_{PQ}} = 4\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359291
When a potential difference of \({10^3}V\) is applied between \(A\) and \(B\), a charge of 0.75 \(mC\) is stored in the system of capacitors as shown. The value of \(C\) is (in \(\mu F\))
1 \(\frac{1}{2}\)
2 \(2\)
3 \(2.5\)
4 \(3\)
Explanation:
The equivalent circuit of the given network is as shown in the figures. Hence, the equivalent capacitance between \(A\) and \(B\) \({C_{eq}}\; = \;\frac{{1\left( {C\; + \;1} \right)}}{{1\; + \;C\; + \;1}}\; = \;\frac{{C\; + \;1}}{{C\; + \;2}}\) The charge stored on the system of capacitors is \(Q = 0.75mC = 0.75 \times {10^{ - 3}}C\) As \(Q = {C_{eq}}V\) \(0.75\; \times \;{10^{ - 3}}\; = \;\frac{{\left( {C\; + \;1} \right)}}{{\left( {C\; + \;2} \right)}}\; \times \;{10^{ - 6}} \times \;{10^3}\) On solving, we get \(C = 2\)
KCET - 2013
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359292
Three capacitors each of \(4\,\mu F\) are to be connected in such a way that the effective capacitance is \(6\,\mu F\). This can be done by connecting them:
1 All in series
2 Two in parallel and one in series
3 Two in series and one in parallel
4 All in parallel
Explanation:
Conceptual Question
JEE - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359293
For the arrangement of capacitors as shown in the circuit, the effective capacitance between the points \(A\) and \(B\) is (capacitance of each capacitor is \(4\mu F\))
1 \(4\mu F\)
2 \(2\mu F\)
3 \(1\mu F\)
4 \(8\mu F\)
Explanation:
The given circuit can be redrawn as shown in figure (ii) As \(\frac{{{C_{AC}}}}{{{C_{AD}}}} = \frac{{{C_{CB}}}}{{{C_{DB}}}}\) The given circuit is a balanced Wheatstone bridge and the capacitance in arm CD is ineffective. Thus, it reduces to the equivalent circuit as shown in figure (iii). As \(4\mu C\) and \(4\mu C\) are in series in the upper branch, their equivalent capacitance \({C_1}\) is \({C_1}\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{4\mu F\; + \;4\mu F}}\; = \;2\mu F\) Similarly, \({4\mu F}\) and \({4\mu F}\) are in series in the lower branch, so their equivalent capacitance \({C_2}\) is \({C_2}\;\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{C\; + \;4\mu F}}\; = \;2\mu F\) And the corresponding equivalent circuit is shown in figure (iv). Since \({C_1}\,{\rm{and}}\,{C_2}\) are in parallel, the effective capacitance between the points A and B is \({C_{AB}} = {C_1} + {C_2} = 2\mu F + 2\mu F = 4\mu F\)
359290
The effective capacitance between the point \(P\) and \(Q\) in the given figure is
1 \(4\;\mu F\)
2 \(16\;\mu F\)
3 \(26\;\mu F\)
4 \(10\;\mu F\)
Explanation:
The points having equal potentials are shown in the figure The circuit can be redrawn as The given circuit is a wheatstone bridge. \({C_{PQ}} = 4\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359291
When a potential difference of \({10^3}V\) is applied between \(A\) and \(B\), a charge of 0.75 \(mC\) is stored in the system of capacitors as shown. The value of \(C\) is (in \(\mu F\))
1 \(\frac{1}{2}\)
2 \(2\)
3 \(2.5\)
4 \(3\)
Explanation:
The equivalent circuit of the given network is as shown in the figures. Hence, the equivalent capacitance between \(A\) and \(B\) \({C_{eq}}\; = \;\frac{{1\left( {C\; + \;1} \right)}}{{1\; + \;C\; + \;1}}\; = \;\frac{{C\; + \;1}}{{C\; + \;2}}\) The charge stored on the system of capacitors is \(Q = 0.75mC = 0.75 \times {10^{ - 3}}C\) As \(Q = {C_{eq}}V\) \(0.75\; \times \;{10^{ - 3}}\; = \;\frac{{\left( {C\; + \;1} \right)}}{{\left( {C\; + \;2} \right)}}\; \times \;{10^{ - 6}} \times \;{10^3}\) On solving, we get \(C = 2\)
KCET - 2013
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359292
Three capacitors each of \(4\,\mu F\) are to be connected in such a way that the effective capacitance is \(6\,\mu F\). This can be done by connecting them:
1 All in series
2 Two in parallel and one in series
3 Two in series and one in parallel
4 All in parallel
Explanation:
Conceptual Question
JEE - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359293
For the arrangement of capacitors as shown in the circuit, the effective capacitance between the points \(A\) and \(B\) is (capacitance of each capacitor is \(4\mu F\))
1 \(4\mu F\)
2 \(2\mu F\)
3 \(1\mu F\)
4 \(8\mu F\)
Explanation:
The given circuit can be redrawn as shown in figure (ii) As \(\frac{{{C_{AC}}}}{{{C_{AD}}}} = \frac{{{C_{CB}}}}{{{C_{DB}}}}\) The given circuit is a balanced Wheatstone bridge and the capacitance in arm CD is ineffective. Thus, it reduces to the equivalent circuit as shown in figure (iii). As \(4\mu C\) and \(4\mu C\) are in series in the upper branch, their equivalent capacitance \({C_1}\) is \({C_1}\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{4\mu F\; + \;4\mu F}}\; = \;2\mu F\) Similarly, \({4\mu F}\) and \({4\mu F}\) are in series in the lower branch, so their equivalent capacitance \({C_2}\) is \({C_2}\;\; = \;\frac{{\left( {4\mu F} \right)\left( {4\mu F} \right)}}{{C\; + \;4\mu F}}\; = \;2\mu F\) And the corresponding equivalent circuit is shown in figure (iv). Since \({C_1}\,{\rm{and}}\,{C_2}\) are in parallel, the effective capacitance between the points A and B is \({C_{AB}} = {C_1} + {C_2} = 2\mu F + 2\mu F = 4\mu F\)
