359303
Assertion : If three capacitors of capacitance \({C_1} < {C_2} < {C_3}\) are connected in parallel then their equivalent capacitance \({C_p} > {C_3}\) Reason : \(\frac{1}{{{C_P}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
359304
Four metal plates are arranged as shown. Capacitance between \(X\) and \(Y\) (\(A\) \( \to \) Area each plate, d \( \to \) distance between the plates)
1 \(\frac{3}{2}\frac{{{\varepsilon _0}A}}{d}\)
2 \(\frac{{2{\varepsilon _0}A}}{d}\)
3 \(\frac{2}{3}\frac{{{\varepsilon _0}A}}{d}\)
4 \(\frac{{3{\varepsilon _0}A}}{d}\)
Explanation:
Given capacitors can be rearranged as Here \(C = \frac{{{\varepsilon _0}A}}{d}\;\;\therefore \;{C_{XY}}\; = \;\frac{2}{3}C\; = \;\frac{2}{3}\;\frac{{{\varepsilon _0}A}}{d}\)
KCET - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359305
The ratio of equivalent capacitance between points \(P\) and \(Q\) in the following cases is
1 \(1: 2: 3\)
2 \(3: 2: 1\)
3 \(1: 1: 1\)
4 \(1: 1: 2\)
Explanation:
The points present on a conducting wire
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359306
The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances \(1\,pF,2pF\,{\rm{and}}\,4pF\) is
1 \(\frac{7}{4}pF\)
2 \(\frac{4}{7}pF\)
3 \(1pF\)
4 \(2pF\)
Explanation:
\({C_1} = 1\,pF,{C_2} = 2pF\,{\rm{and}}\,{C_3} = 4pF\) Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance \(C\)' will be, \(\frac{1}{{C'}}\; = \;1 + \;\frac{1}{2}\; + \;\frac{1}{4}\; = \,\frac{7}{4}\;\;\;{\rm{or}}\;\;\;C'\; = \;\frac{4}{7}pF\)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359303
Assertion : If three capacitors of capacitance \({C_1} < {C_2} < {C_3}\) are connected in parallel then their equivalent capacitance \({C_p} > {C_3}\) Reason : \(\frac{1}{{{C_P}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
359304
Four metal plates are arranged as shown. Capacitance between \(X\) and \(Y\) (\(A\) \( \to \) Area each plate, d \( \to \) distance between the plates)
1 \(\frac{3}{2}\frac{{{\varepsilon _0}A}}{d}\)
2 \(\frac{{2{\varepsilon _0}A}}{d}\)
3 \(\frac{2}{3}\frac{{{\varepsilon _0}A}}{d}\)
4 \(\frac{{3{\varepsilon _0}A}}{d}\)
Explanation:
Given capacitors can be rearranged as Here \(C = \frac{{{\varepsilon _0}A}}{d}\;\;\therefore \;{C_{XY}}\; = \;\frac{2}{3}C\; = \;\frac{2}{3}\;\frac{{{\varepsilon _0}A}}{d}\)
KCET - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359305
The ratio of equivalent capacitance between points \(P\) and \(Q\) in the following cases is
1 \(1: 2: 3\)
2 \(3: 2: 1\)
3 \(1: 1: 1\)
4 \(1: 1: 2\)
Explanation:
The points present on a conducting wire
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359306
The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances \(1\,pF,2pF\,{\rm{and}}\,4pF\) is
1 \(\frac{7}{4}pF\)
2 \(\frac{4}{7}pF\)
3 \(1pF\)
4 \(2pF\)
Explanation:
\({C_1} = 1\,pF,{C_2} = 2pF\,{\rm{and}}\,{C_3} = 4pF\) Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance \(C\)' will be, \(\frac{1}{{C'}}\; = \;1 + \;\frac{1}{2}\; + \;\frac{1}{4}\; = \,\frac{7}{4}\;\;\;{\rm{or}}\;\;\;C'\; = \;\frac{4}{7}pF\)
359303
Assertion : If three capacitors of capacitance \({C_1} < {C_2} < {C_3}\) are connected in parallel then their equivalent capacitance \({C_p} > {C_3}\) Reason : \(\frac{1}{{{C_P}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
359304
Four metal plates are arranged as shown. Capacitance between \(X\) and \(Y\) (\(A\) \( \to \) Area each plate, d \( \to \) distance between the plates)
1 \(\frac{3}{2}\frac{{{\varepsilon _0}A}}{d}\)
2 \(\frac{{2{\varepsilon _0}A}}{d}\)
3 \(\frac{2}{3}\frac{{{\varepsilon _0}A}}{d}\)
4 \(\frac{{3{\varepsilon _0}A}}{d}\)
Explanation:
Given capacitors can be rearranged as Here \(C = \frac{{{\varepsilon _0}A}}{d}\;\;\therefore \;{C_{XY}}\; = \;\frac{2}{3}C\; = \;\frac{2}{3}\;\frac{{{\varepsilon _0}A}}{d}\)
KCET - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359305
The ratio of equivalent capacitance between points \(P\) and \(Q\) in the following cases is
1 \(1: 2: 3\)
2 \(3: 2: 1\)
3 \(1: 1: 1\)
4 \(1: 1: 2\)
Explanation:
The points present on a conducting wire
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359306
The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances \(1\,pF,2pF\,{\rm{and}}\,4pF\) is
1 \(\frac{7}{4}pF\)
2 \(\frac{4}{7}pF\)
3 \(1pF\)
4 \(2pF\)
Explanation:
\({C_1} = 1\,pF,{C_2} = 2pF\,{\rm{and}}\,{C_3} = 4pF\) Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance \(C\)' will be, \(\frac{1}{{C'}}\; = \;1 + \;\frac{1}{2}\; + \;\frac{1}{4}\; = \,\frac{7}{4}\;\;\;{\rm{or}}\;\;\;C'\; = \;\frac{4}{7}pF\)
359303
Assertion : If three capacitors of capacitance \({C_1} < {C_2} < {C_3}\) are connected in parallel then their equivalent capacitance \({C_p} > {C_3}\) Reason : \(\frac{1}{{{C_P}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
359304
Four metal plates are arranged as shown. Capacitance between \(X\) and \(Y\) (\(A\) \( \to \) Area each plate, d \( \to \) distance between the plates)
1 \(\frac{3}{2}\frac{{{\varepsilon _0}A}}{d}\)
2 \(\frac{{2{\varepsilon _0}A}}{d}\)
3 \(\frac{2}{3}\frac{{{\varepsilon _0}A}}{d}\)
4 \(\frac{{3{\varepsilon _0}A}}{d}\)
Explanation:
Given capacitors can be rearranged as Here \(C = \frac{{{\varepsilon _0}A}}{d}\;\;\therefore \;{C_{XY}}\; = \;\frac{2}{3}C\; = \;\frac{2}{3}\;\frac{{{\varepsilon _0}A}}{d}\)
KCET - 2016
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359305
The ratio of equivalent capacitance between points \(P\) and \(Q\) in the following cases is
1 \(1: 2: 3\)
2 \(3: 2: 1\)
3 \(1: 1: 1\)
4 \(1: 1: 2\)
Explanation:
The points present on a conducting wire
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359306
The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances \(1\,pF,2pF\,{\rm{and}}\,4pF\) is
1 \(\frac{7}{4}pF\)
2 \(\frac{4}{7}pF\)
3 \(1pF\)
4 \(2pF\)
Explanation:
\({C_1} = 1\,pF,{C_2} = 2pF\,{\rm{and}}\,{C_3} = 4pF\) Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance \(C\)' will be, \(\frac{1}{{C'}}\; = \;1 + \;\frac{1}{2}\; + \;\frac{1}{4}\; = \,\frac{7}{4}\;\;\;{\rm{or}}\;\;\;C'\; = \;\frac{4}{7}pF\)
