359298
Three capacitors each of capacitance \(C\) and breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be
1 \(3C,V/3\)
2 \(C/3,3V\)
3 \(3C,3V\)
4 \(C/3,V/3\)
Explanation:
\({C_{eq}} = \frac{C}{3}{V_{Max}} = V + V + V = 3V\)
NCERT Exemplar
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359299
A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 \(cm\) and the overlapping area of the plates is \(5\,c{m^2}\). The capacity of the unit is
1 \(1.06\,pF\)
2 \(4\,pF\)
3 \(6.36\,pF\)
4 \(12.72\,pF\)
Explanation:
The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, \(C\; = \;\frac{{{\varepsilon _0}A}}{d}\; = \,\frac{{8.854\; \times \;{{10}^{ - 12}}\; \times \;5\; \times \;{{10}^{ - 4}}}}{{0.885\; \times \;{{10}^{ - 2}}}}\; = \;0.5pF\) The total capacity of 8 capacitors \( = 8C\) \( = 8 \times 0.5 = 4pF\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359300
The equivalent capacitance between \(A\) and \(B\) in the circuit given below is:
1 \(3.6\,\mu F\)
2 \(2.4\,\mu F\)
3 \(4.9\,\mu F\)
4 \(5.4\,\mu F\)
Explanation:
By joining the points having identical potentials \(\frac{1}{{{C_{eq}}}} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{6} = \frac{5}{{12}}\; \Rightarrow \,\;{C_{eq}} = \frac{{12}}{5} = 2.4\,\mu F\)
JEE - 2018
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359301
For circuit, the equivalent capacitance between \(P\) and \(Q\) is
359302
The equivalent capacitance of the system shown in the following circuit is:
1 \(3 \mu F\)
2 \(6 \mu F\)
3 \(9 \mu F\)
4 \(2 \mu F\)
Explanation:
The two capacitors on right side are in parallel connection equivalent to \((3+3)=6 \mu F\). Then, that \(6 \mu F\) is in series with \(3 \mu F\) of left side. \( \Rightarrow {C_{AB}} = \frac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}\) \( \Rightarrow {C_{AB}} = \frac{{3 \times 6}}{{3 + 6}} = \frac{{18}}{9} = 2\mu F\) Correct option is (4).
359298
Three capacitors each of capacitance \(C\) and breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be
1 \(3C,V/3\)
2 \(C/3,3V\)
3 \(3C,3V\)
4 \(C/3,V/3\)
Explanation:
\({C_{eq}} = \frac{C}{3}{V_{Max}} = V + V + V = 3V\)
NCERT Exemplar
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359299
A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 \(cm\) and the overlapping area of the plates is \(5\,c{m^2}\). The capacity of the unit is
1 \(1.06\,pF\)
2 \(4\,pF\)
3 \(6.36\,pF\)
4 \(12.72\,pF\)
Explanation:
The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, \(C\; = \;\frac{{{\varepsilon _0}A}}{d}\; = \,\frac{{8.854\; \times \;{{10}^{ - 12}}\; \times \;5\; \times \;{{10}^{ - 4}}}}{{0.885\; \times \;{{10}^{ - 2}}}}\; = \;0.5pF\) The total capacity of 8 capacitors \( = 8C\) \( = 8 \times 0.5 = 4pF\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359300
The equivalent capacitance between \(A\) and \(B\) in the circuit given below is:
1 \(3.6\,\mu F\)
2 \(2.4\,\mu F\)
3 \(4.9\,\mu F\)
4 \(5.4\,\mu F\)
Explanation:
By joining the points having identical potentials \(\frac{1}{{{C_{eq}}}} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{6} = \frac{5}{{12}}\; \Rightarrow \,\;{C_{eq}} = \frac{{12}}{5} = 2.4\,\mu F\)
JEE - 2018
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359301
For circuit, the equivalent capacitance between \(P\) and \(Q\) is
359302
The equivalent capacitance of the system shown in the following circuit is:
1 \(3 \mu F\)
2 \(6 \mu F\)
3 \(9 \mu F\)
4 \(2 \mu F\)
Explanation:
The two capacitors on right side are in parallel connection equivalent to \((3+3)=6 \mu F\). Then, that \(6 \mu F\) is in series with \(3 \mu F\) of left side. \( \Rightarrow {C_{AB}} = \frac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}\) \( \Rightarrow {C_{AB}} = \frac{{3 \times 6}}{{3 + 6}} = \frac{{18}}{9} = 2\mu F\) Correct option is (4).
359298
Three capacitors each of capacitance \(C\) and breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be
1 \(3C,V/3\)
2 \(C/3,3V\)
3 \(3C,3V\)
4 \(C/3,V/3\)
Explanation:
\({C_{eq}} = \frac{C}{3}{V_{Max}} = V + V + V = 3V\)
NCERT Exemplar
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359299
A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 \(cm\) and the overlapping area of the plates is \(5\,c{m^2}\). The capacity of the unit is
1 \(1.06\,pF\)
2 \(4\,pF\)
3 \(6.36\,pF\)
4 \(12.72\,pF\)
Explanation:
The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, \(C\; = \;\frac{{{\varepsilon _0}A}}{d}\; = \,\frac{{8.854\; \times \;{{10}^{ - 12}}\; \times \;5\; \times \;{{10}^{ - 4}}}}{{0.885\; \times \;{{10}^{ - 2}}}}\; = \;0.5pF\) The total capacity of 8 capacitors \( = 8C\) \( = 8 \times 0.5 = 4pF\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359300
The equivalent capacitance between \(A\) and \(B\) in the circuit given below is:
1 \(3.6\,\mu F\)
2 \(2.4\,\mu F\)
3 \(4.9\,\mu F\)
4 \(5.4\,\mu F\)
Explanation:
By joining the points having identical potentials \(\frac{1}{{{C_{eq}}}} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{6} = \frac{5}{{12}}\; \Rightarrow \,\;{C_{eq}} = \frac{{12}}{5} = 2.4\,\mu F\)
JEE - 2018
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359301
For circuit, the equivalent capacitance between \(P\) and \(Q\) is
359302
The equivalent capacitance of the system shown in the following circuit is:
1 \(3 \mu F\)
2 \(6 \mu F\)
3 \(9 \mu F\)
4 \(2 \mu F\)
Explanation:
The two capacitors on right side are in parallel connection equivalent to \((3+3)=6 \mu F\). Then, that \(6 \mu F\) is in series with \(3 \mu F\) of left side. \( \Rightarrow {C_{AB}} = \frac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}\) \( \Rightarrow {C_{AB}} = \frac{{3 \times 6}}{{3 + 6}} = \frac{{18}}{9} = 2\mu F\) Correct option is (4).
359298
Three capacitors each of capacitance \(C\) and breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be
1 \(3C,V/3\)
2 \(C/3,3V\)
3 \(3C,3V\)
4 \(C/3,V/3\)
Explanation:
\({C_{eq}} = \frac{C}{3}{V_{Max}} = V + V + V = 3V\)
NCERT Exemplar
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359299
A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 \(cm\) and the overlapping area of the plates is \(5\,c{m^2}\). The capacity of the unit is
1 \(1.06\,pF\)
2 \(4\,pF\)
3 \(6.36\,pF\)
4 \(12.72\,pF\)
Explanation:
The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, \(C\; = \;\frac{{{\varepsilon _0}A}}{d}\; = \,\frac{{8.854\; \times \;{{10}^{ - 12}}\; \times \;5\; \times \;{{10}^{ - 4}}}}{{0.885\; \times \;{{10}^{ - 2}}}}\; = \;0.5pF\) The total capacity of 8 capacitors \( = 8C\) \( = 8 \times 0.5 = 4pF\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359300
The equivalent capacitance between \(A\) and \(B\) in the circuit given below is:
1 \(3.6\,\mu F\)
2 \(2.4\,\mu F\)
3 \(4.9\,\mu F\)
4 \(5.4\,\mu F\)
Explanation:
By joining the points having identical potentials \(\frac{1}{{{C_{eq}}}} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{6} = \frac{5}{{12}}\; \Rightarrow \,\;{C_{eq}} = \frac{{12}}{5} = 2.4\,\mu F\)
JEE - 2018
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359301
For circuit, the equivalent capacitance between \(P\) and \(Q\) is
359302
The equivalent capacitance of the system shown in the following circuit is:
1 \(3 \mu F\)
2 \(6 \mu F\)
3 \(9 \mu F\)
4 \(2 \mu F\)
Explanation:
The two capacitors on right side are in parallel connection equivalent to \((3+3)=6 \mu F\). Then, that \(6 \mu F\) is in series with \(3 \mu F\) of left side. \( \Rightarrow {C_{AB}} = \frac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}\) \( \Rightarrow {C_{AB}} = \frac{{3 \times 6}}{{3 + 6}} = \frac{{18}}{9} = 2\mu F\) Correct option is (4).
359298
Three capacitors each of capacitance \(C\) and breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be
1 \(3C,V/3\)
2 \(C/3,3V\)
3 \(3C,3V\)
4 \(C/3,V/3\)
Explanation:
\({C_{eq}} = \frac{C}{3}{V_{Max}} = V + V + V = 3V\)
NCERT Exemplar
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359299
A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 \(cm\) and the overlapping area of the plates is \(5\,c{m^2}\). The capacity of the unit is
1 \(1.06\,pF\)
2 \(4\,pF\)
3 \(6.36\,pF\)
4 \(12.72\,pF\)
Explanation:
The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, \(C\; = \;\frac{{{\varepsilon _0}A}}{d}\; = \,\frac{{8.854\; \times \;{{10}^{ - 12}}\; \times \;5\; \times \;{{10}^{ - 4}}}}{{0.885\; \times \;{{10}^{ - 2}}}}\; = \;0.5pF\) The total capacity of 8 capacitors \( = 8C\) \( = 8 \times 0.5 = 4pF\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359300
The equivalent capacitance between \(A\) and \(B\) in the circuit given below is:
1 \(3.6\,\mu F\)
2 \(2.4\,\mu F\)
3 \(4.9\,\mu F\)
4 \(5.4\,\mu F\)
Explanation:
By joining the points having identical potentials \(\frac{1}{{{C_{eq}}}} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{6} = \frac{5}{{12}}\; \Rightarrow \,\;{C_{eq}} = \frac{{12}}{5} = 2.4\,\mu F\)
JEE - 2018
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359301
For circuit, the equivalent capacitance between \(P\) and \(Q\) is
359302
The equivalent capacitance of the system shown in the following circuit is:
1 \(3 \mu F\)
2 \(6 \mu F\)
3 \(9 \mu F\)
4 \(2 \mu F\)
Explanation:
The two capacitors on right side are in parallel connection equivalent to \((3+3)=6 \mu F\). Then, that \(6 \mu F\) is in series with \(3 \mu F\) of left side. \( \Rightarrow {C_{AB}} = \frac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}\) \( \Rightarrow {C_{AB}} = \frac{{3 \times 6}}{{3 + 6}} = \frac{{18}}{9} = 2\mu F\) Correct option is (4).
