359256
The expression for the equivalent capacitance of the system shown in the figure is (\(A\) is the cross - sectional area of one of the plates)
1 \(\frac{{{\varepsilon _o}A}}{{6d}}\)
2 \(\frac{{{\varepsilon _o}A}}{{3d}}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{11{\varepsilon _o}A}}{{6d}}\)
Explanation:
The given system can be assumed as a combination of three capacitors connected in parallel, as shown in the figure \( \Rightarrow {C_{eq}} = {C_1} + {C_2} + {C_3}\) \({C_{eq}} = \frac{{{\varepsilon _o}A}}{d} + \frac{{{\varepsilon _o}A}}{{2d}} + \frac{{{\varepsilon _o}A}}{{3d}} = \frac{{11{\varepsilon _o}A}}{{6d}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359257
A capacitor is made of a flat plate of area \(A\) and a second plate having a stair-like structure as shown in figure. If the area of each stair is \(\dfrac{A}{3}\) and the height is \(d\), the capacitance of the arrangement is
1 \(\dfrac{11 \varepsilon_{0} A}{20 d}\)
2 \(\dfrac{18 \varepsilon_{0} A}{11 d}\)
3 \(\dfrac{11 \varepsilon_{0} A}{18 d}\)
4 \(\dfrac{13 \varepsilon_{0} A}{17 d}\)
Explanation:
Capacitance for \({I^{st}}\,\,{\text{stair}},{\text{ }}{C_1} = \frac{{{\varepsilon _0}A/3}}{d} = \frac{{{\varepsilon _0}A}}{{3d}}\) Capacitance for \(I{I^{nd{\text{ }}}}\) stair, \(C_{2}=\dfrac{\varepsilon_{0} A / 3}{2 d}=\dfrac{\varepsilon_{0} A}{6 d}\) Capacitance for \(II{I^{rd}}\) stair, \(C_{3}=\dfrac{\varepsilon_{0} A / 3}{3 d}=\dfrac{\varepsilon_{0} A}{9 d}\) As, \(C_{1}, C_{2}\) and \(C_{3}\) are in parallel, \(\therefore {C_{eq}} = {C_1} + {C_2} + {C_3} = \frac{{{\varepsilon _0}A}}{{3d}}\left( {1 + \frac{1}{2} + \frac{1}{3}} \right)\) \( \Rightarrow {C_{eq}} = \frac{{11\,{\varepsilon _0}A}}{{18\,d}}\)
JEE - 2024
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359258
When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is \(3.75\mu F\). The capacity of each capacitor is
1 \(5\mu F\)
2 \(6\mu F\)
3 \(7\mu F\)
4 \(8\mu F\)
Explanation:
When three identical capacitors, each of capacitance \(C\), is connected in parallel, the total capacitance of the combination \(C' = C + C + C = 3C\) When a fourth capacitor of capacitance \(C\) is connected in series with this combination, the total equivalent capacitance is \({C_{eq}} = \frac{{3C \times C}}{{3C + C}}\) \( \Rightarrow {C_{eq}} = \frac{{3C}}{4}\) \( \Rightarrow C = \frac{{4{C_{eq}}}}{3} = \frac{{4 \times 3.75}}{3} = 5.00\mu F\)
MHTCET - 2017
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359259
Five capacitors each of value \(1 \mu F\) are connected as shown in the figure. The equivalent capacitance between \(A\) and \(B\) is
1 \(2 \mu F\)
2 \(5 \mu F\)
3 \(3 \mu F\)
4 \(1 \mu F\)
Explanation:
Each pair (left, right) is in series connection. They are in ‘parallel-connection’. \( \Rightarrow {C_{AB}} = {C_1} + {C_2}\) \( \Rightarrow {C_{AB}} = \frac{1}{2} + \frac{1}{2}\) \({C_{AB}} = 1\mu F\) Correct option is (4).
359256
The expression for the equivalent capacitance of the system shown in the figure is (\(A\) is the cross - sectional area of one of the plates)
1 \(\frac{{{\varepsilon _o}A}}{{6d}}\)
2 \(\frac{{{\varepsilon _o}A}}{{3d}}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{11{\varepsilon _o}A}}{{6d}}\)
Explanation:
The given system can be assumed as a combination of three capacitors connected in parallel, as shown in the figure \( \Rightarrow {C_{eq}} = {C_1} + {C_2} + {C_3}\) \({C_{eq}} = \frac{{{\varepsilon _o}A}}{d} + \frac{{{\varepsilon _o}A}}{{2d}} + \frac{{{\varepsilon _o}A}}{{3d}} = \frac{{11{\varepsilon _o}A}}{{6d}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359257
A capacitor is made of a flat plate of area \(A\) and a second plate having a stair-like structure as shown in figure. If the area of each stair is \(\dfrac{A}{3}\) and the height is \(d\), the capacitance of the arrangement is
1 \(\dfrac{11 \varepsilon_{0} A}{20 d}\)
2 \(\dfrac{18 \varepsilon_{0} A}{11 d}\)
3 \(\dfrac{11 \varepsilon_{0} A}{18 d}\)
4 \(\dfrac{13 \varepsilon_{0} A}{17 d}\)
Explanation:
Capacitance for \({I^{st}}\,\,{\text{stair}},{\text{ }}{C_1} = \frac{{{\varepsilon _0}A/3}}{d} = \frac{{{\varepsilon _0}A}}{{3d}}\) Capacitance for \(I{I^{nd{\text{ }}}}\) stair, \(C_{2}=\dfrac{\varepsilon_{0} A / 3}{2 d}=\dfrac{\varepsilon_{0} A}{6 d}\) Capacitance for \(II{I^{rd}}\) stair, \(C_{3}=\dfrac{\varepsilon_{0} A / 3}{3 d}=\dfrac{\varepsilon_{0} A}{9 d}\) As, \(C_{1}, C_{2}\) and \(C_{3}\) are in parallel, \(\therefore {C_{eq}} = {C_1} + {C_2} + {C_3} = \frac{{{\varepsilon _0}A}}{{3d}}\left( {1 + \frac{1}{2} + \frac{1}{3}} \right)\) \( \Rightarrow {C_{eq}} = \frac{{11\,{\varepsilon _0}A}}{{18\,d}}\)
JEE - 2024
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359258
When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is \(3.75\mu F\). The capacity of each capacitor is
1 \(5\mu F\)
2 \(6\mu F\)
3 \(7\mu F\)
4 \(8\mu F\)
Explanation:
When three identical capacitors, each of capacitance \(C\), is connected in parallel, the total capacitance of the combination \(C' = C + C + C = 3C\) When a fourth capacitor of capacitance \(C\) is connected in series with this combination, the total equivalent capacitance is \({C_{eq}} = \frac{{3C \times C}}{{3C + C}}\) \( \Rightarrow {C_{eq}} = \frac{{3C}}{4}\) \( \Rightarrow C = \frac{{4{C_{eq}}}}{3} = \frac{{4 \times 3.75}}{3} = 5.00\mu F\)
MHTCET - 2017
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359259
Five capacitors each of value \(1 \mu F\) are connected as shown in the figure. The equivalent capacitance between \(A\) and \(B\) is
1 \(2 \mu F\)
2 \(5 \mu F\)
3 \(3 \mu F\)
4 \(1 \mu F\)
Explanation:
Each pair (left, right) is in series connection. They are in ‘parallel-connection’. \( \Rightarrow {C_{AB}} = {C_1} + {C_2}\) \( \Rightarrow {C_{AB}} = \frac{1}{2} + \frac{1}{2}\) \({C_{AB}} = 1\mu F\) Correct option is (4).
359256
The expression for the equivalent capacitance of the system shown in the figure is (\(A\) is the cross - sectional area of one of the plates)
1 \(\frac{{{\varepsilon _o}A}}{{6d}}\)
2 \(\frac{{{\varepsilon _o}A}}{{3d}}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{11{\varepsilon _o}A}}{{6d}}\)
Explanation:
The given system can be assumed as a combination of three capacitors connected in parallel, as shown in the figure \( \Rightarrow {C_{eq}} = {C_1} + {C_2} + {C_3}\) \({C_{eq}} = \frac{{{\varepsilon _o}A}}{d} + \frac{{{\varepsilon _o}A}}{{2d}} + \frac{{{\varepsilon _o}A}}{{3d}} = \frac{{11{\varepsilon _o}A}}{{6d}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359257
A capacitor is made of a flat plate of area \(A\) and a second plate having a stair-like structure as shown in figure. If the area of each stair is \(\dfrac{A}{3}\) and the height is \(d\), the capacitance of the arrangement is
1 \(\dfrac{11 \varepsilon_{0} A}{20 d}\)
2 \(\dfrac{18 \varepsilon_{0} A}{11 d}\)
3 \(\dfrac{11 \varepsilon_{0} A}{18 d}\)
4 \(\dfrac{13 \varepsilon_{0} A}{17 d}\)
Explanation:
Capacitance for \({I^{st}}\,\,{\text{stair}},{\text{ }}{C_1} = \frac{{{\varepsilon _0}A/3}}{d} = \frac{{{\varepsilon _0}A}}{{3d}}\) Capacitance for \(I{I^{nd{\text{ }}}}\) stair, \(C_{2}=\dfrac{\varepsilon_{0} A / 3}{2 d}=\dfrac{\varepsilon_{0} A}{6 d}\) Capacitance for \(II{I^{rd}}\) stair, \(C_{3}=\dfrac{\varepsilon_{0} A / 3}{3 d}=\dfrac{\varepsilon_{0} A}{9 d}\) As, \(C_{1}, C_{2}\) and \(C_{3}\) are in parallel, \(\therefore {C_{eq}} = {C_1} + {C_2} + {C_3} = \frac{{{\varepsilon _0}A}}{{3d}}\left( {1 + \frac{1}{2} + \frac{1}{3}} \right)\) \( \Rightarrow {C_{eq}} = \frac{{11\,{\varepsilon _0}A}}{{18\,d}}\)
JEE - 2024
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359258
When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is \(3.75\mu F\). The capacity of each capacitor is
1 \(5\mu F\)
2 \(6\mu F\)
3 \(7\mu F\)
4 \(8\mu F\)
Explanation:
When three identical capacitors, each of capacitance \(C\), is connected in parallel, the total capacitance of the combination \(C' = C + C + C = 3C\) When a fourth capacitor of capacitance \(C\) is connected in series with this combination, the total equivalent capacitance is \({C_{eq}} = \frac{{3C \times C}}{{3C + C}}\) \( \Rightarrow {C_{eq}} = \frac{{3C}}{4}\) \( \Rightarrow C = \frac{{4{C_{eq}}}}{3} = \frac{{4 \times 3.75}}{3} = 5.00\mu F\)
MHTCET - 2017
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359259
Five capacitors each of value \(1 \mu F\) are connected as shown in the figure. The equivalent capacitance between \(A\) and \(B\) is
1 \(2 \mu F\)
2 \(5 \mu F\)
3 \(3 \mu F\)
4 \(1 \mu F\)
Explanation:
Each pair (left, right) is in series connection. They are in ‘parallel-connection’. \( \Rightarrow {C_{AB}} = {C_1} + {C_2}\) \( \Rightarrow {C_{AB}} = \frac{1}{2} + \frac{1}{2}\) \({C_{AB}} = 1\mu F\) Correct option is (4).
359256
The expression for the equivalent capacitance of the system shown in the figure is (\(A\) is the cross - sectional area of one of the plates)
1 \(\frac{{{\varepsilon _o}A}}{{6d}}\)
2 \(\frac{{{\varepsilon _o}A}}{{3d}}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{11{\varepsilon _o}A}}{{6d}}\)
Explanation:
The given system can be assumed as a combination of three capacitors connected in parallel, as shown in the figure \( \Rightarrow {C_{eq}} = {C_1} + {C_2} + {C_3}\) \({C_{eq}} = \frac{{{\varepsilon _o}A}}{d} + \frac{{{\varepsilon _o}A}}{{2d}} + \frac{{{\varepsilon _o}A}}{{3d}} = \frac{{11{\varepsilon _o}A}}{{6d}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359257
A capacitor is made of a flat plate of area \(A\) and a second plate having a stair-like structure as shown in figure. If the area of each stair is \(\dfrac{A}{3}\) and the height is \(d\), the capacitance of the arrangement is
1 \(\dfrac{11 \varepsilon_{0} A}{20 d}\)
2 \(\dfrac{18 \varepsilon_{0} A}{11 d}\)
3 \(\dfrac{11 \varepsilon_{0} A}{18 d}\)
4 \(\dfrac{13 \varepsilon_{0} A}{17 d}\)
Explanation:
Capacitance for \({I^{st}}\,\,{\text{stair}},{\text{ }}{C_1} = \frac{{{\varepsilon _0}A/3}}{d} = \frac{{{\varepsilon _0}A}}{{3d}}\) Capacitance for \(I{I^{nd{\text{ }}}}\) stair, \(C_{2}=\dfrac{\varepsilon_{0} A / 3}{2 d}=\dfrac{\varepsilon_{0} A}{6 d}\) Capacitance for \(II{I^{rd}}\) stair, \(C_{3}=\dfrac{\varepsilon_{0} A / 3}{3 d}=\dfrac{\varepsilon_{0} A}{9 d}\) As, \(C_{1}, C_{2}\) and \(C_{3}\) are in parallel, \(\therefore {C_{eq}} = {C_1} + {C_2} + {C_3} = \frac{{{\varepsilon _0}A}}{{3d}}\left( {1 + \frac{1}{2} + \frac{1}{3}} \right)\) \( \Rightarrow {C_{eq}} = \frac{{11\,{\varepsilon _0}A}}{{18\,d}}\)
JEE - 2024
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359258
When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is \(3.75\mu F\). The capacity of each capacitor is
1 \(5\mu F\)
2 \(6\mu F\)
3 \(7\mu F\)
4 \(8\mu F\)
Explanation:
When three identical capacitors, each of capacitance \(C\), is connected in parallel, the total capacitance of the combination \(C' = C + C + C = 3C\) When a fourth capacitor of capacitance \(C\) is connected in series with this combination, the total equivalent capacitance is \({C_{eq}} = \frac{{3C \times C}}{{3C + C}}\) \( \Rightarrow {C_{eq}} = \frac{{3C}}{4}\) \( \Rightarrow C = \frac{{4{C_{eq}}}}{3} = \frac{{4 \times 3.75}}{3} = 5.00\mu F\)
MHTCET - 2017
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359259
Five capacitors each of value \(1 \mu F\) are connected as shown in the figure. The equivalent capacitance between \(A\) and \(B\) is
1 \(2 \mu F\)
2 \(5 \mu F\)
3 \(3 \mu F\)
4 \(1 \mu F\)
Explanation:
Each pair (left, right) is in series connection. They are in ‘parallel-connection’. \( \Rightarrow {C_{AB}} = {C_1} + {C_2}\) \( \Rightarrow {C_{AB}} = \frac{1}{2} + \frac{1}{2}\) \({C_{AB}} = 1\mu F\) Correct option is (4).
