CAPACITORSIN SERIES AND IN PARALLEL
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Electrostatic Potentials and Capacitance

268103 If \(\mathbf{3}\) capacitors of values 1,2 and \(3 \mu F\) are available. Themaximum and minimum values of capacitances one can obtain by different combinations of the three capacitors together are respectively .... and

1 \(6 \mu F, \frac{6}{11} \mu F\)
2 \(6 \mu F, \frac{11}{6} \mu F\)
3 \(3 \mu F, 1 \mu F\)
4 \(4 \mu F, 2 \mu F\)
Electrostatic Potentials and Capacitance

268111 Two condensers of capacity \(\mathrm{C}\) and \(2 \mathrm{C}\) are connected in parallel and these are charged upto \(\mathrm{V}\) volt. If the battery is removed and dielectric medium of constant \(K\) is put between the plates of first condenser, then the potential at each condenser is

1 \(\frac{V}{k+2}\)
2 \(2+\frac{k}{3 V}\)
3 \(\frac{2 V}{k+2}\)
4 \(\frac{3 V}{k+2}\)
Electrostatic Potentials and Capacitance

268112 Given a number of capacitors labelled as \(\mathrm{C}, \mathrm{V}\). Find the minimum number of capacitors needed to get an arrangement equivalent to \(C_{\text {net }}, V_{\text {net }}\)

1 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}{ }^{2}}{V^{2}}\)
2 \(n=\frac{C}{C_{\text {net }}} \times \frac{V^{2}}{V_{\text {net }}{ }^{2}}\)
3 \(n=\frac{C}{C_{n e t}} \times \frac{V}{V_{n e t}}\)
4 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}}{V}\)
Electrostatic Potentials and Capacitance

268113 Two capacitors of capacities \(3 \mu F\) and \(6 \mu F\) are connected in series and connected to 120V. The potential difference across \(3 \mu \mathrm{F}\) is \(V_{0}\) and the charge here is \(q_{0}\). We have
A) \(q_{0}=40 \mu C\)
B) \(V_{0}=60 \mathrm{~V}\)
C) \(V_{0}=80 \mathrm{~V}\)
D) \(q_{0}=240 \mu C\)

1 A, C are correct
2 A, B are correct
3 B, D are correct
4 C, D are correct
Electrostatic Potentials and Capacitance

268114 \(\mathbf{n}\) C apacitors of \(2 \mu \mathrm{F}\) each are connected in parallel and a p.d of \(200 \mathrm{v}\) is applied to the combination. The total charge on them was \(1 c\) then \(n\) is equal to

1 3333
2 3000
3 2500
4 25
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Electrostatic Potentials and Capacitance

268103 If \(\mathbf{3}\) capacitors of values 1,2 and \(3 \mu F\) are available. Themaximum and minimum values of capacitances one can obtain by different combinations of the three capacitors together are respectively .... and

1 \(6 \mu F, \frac{6}{11} \mu F\)
2 \(6 \mu F, \frac{11}{6} \mu F\)
3 \(3 \mu F, 1 \mu F\)
4 \(4 \mu F, 2 \mu F\)
Electrostatic Potentials and Capacitance

268111 Two condensers of capacity \(\mathrm{C}\) and \(2 \mathrm{C}\) are connected in parallel and these are charged upto \(\mathrm{V}\) volt. If the battery is removed and dielectric medium of constant \(K\) is put between the plates of first condenser, then the potential at each condenser is

1 \(\frac{V}{k+2}\)
2 \(2+\frac{k}{3 V}\)
3 \(\frac{2 V}{k+2}\)
4 \(\frac{3 V}{k+2}\)
Electrostatic Potentials and Capacitance

268112 Given a number of capacitors labelled as \(\mathrm{C}, \mathrm{V}\). Find the minimum number of capacitors needed to get an arrangement equivalent to \(C_{\text {net }}, V_{\text {net }}\)

1 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}{ }^{2}}{V^{2}}\)
2 \(n=\frac{C}{C_{\text {net }}} \times \frac{V^{2}}{V_{\text {net }}{ }^{2}}\)
3 \(n=\frac{C}{C_{n e t}} \times \frac{V}{V_{n e t}}\)
4 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}}{V}\)
Electrostatic Potentials and Capacitance

268113 Two capacitors of capacities \(3 \mu F\) and \(6 \mu F\) are connected in series and connected to 120V. The potential difference across \(3 \mu \mathrm{F}\) is \(V_{0}\) and the charge here is \(q_{0}\). We have
A) \(q_{0}=40 \mu C\)
B) \(V_{0}=60 \mathrm{~V}\)
C) \(V_{0}=80 \mathrm{~V}\)
D) \(q_{0}=240 \mu C\)

1 A, C are correct
2 A, B are correct
3 B, D are correct
4 C, D are correct
Electrostatic Potentials and Capacitance

268114 \(\mathbf{n}\) C apacitors of \(2 \mu \mathrm{F}\) each are connected in parallel and a p.d of \(200 \mathrm{v}\) is applied to the combination. The total charge on them was \(1 c\) then \(n\) is equal to

1 3333
2 3000
3 2500
4 25
For More Updates join with Us
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Electrostatic Potentials and Capacitance

268103 If \(\mathbf{3}\) capacitors of values 1,2 and \(3 \mu F\) are available. Themaximum and minimum values of capacitances one can obtain by different combinations of the three capacitors together are respectively .... and

1 \(6 \mu F, \frac{6}{11} \mu F\)
2 \(6 \mu F, \frac{11}{6} \mu F\)
3 \(3 \mu F, 1 \mu F\)
4 \(4 \mu F, 2 \mu F\)
Electrostatic Potentials and Capacitance

268111 Two condensers of capacity \(\mathrm{C}\) and \(2 \mathrm{C}\) are connected in parallel and these are charged upto \(\mathrm{V}\) volt. If the battery is removed and dielectric medium of constant \(K\) is put between the plates of first condenser, then the potential at each condenser is

1 \(\frac{V}{k+2}\)
2 \(2+\frac{k}{3 V}\)
3 \(\frac{2 V}{k+2}\)
4 \(\frac{3 V}{k+2}\)
Electrostatic Potentials and Capacitance

268112 Given a number of capacitors labelled as \(\mathrm{C}, \mathrm{V}\). Find the minimum number of capacitors needed to get an arrangement equivalent to \(C_{\text {net }}, V_{\text {net }}\)

1 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}{ }^{2}}{V^{2}}\)
2 \(n=\frac{C}{C_{\text {net }}} \times \frac{V^{2}}{V_{\text {net }}{ }^{2}}\)
3 \(n=\frac{C}{C_{n e t}} \times \frac{V}{V_{n e t}}\)
4 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}}{V}\)
Electrostatic Potentials and Capacitance

268113 Two capacitors of capacities \(3 \mu F\) and \(6 \mu F\) are connected in series and connected to 120V. The potential difference across \(3 \mu \mathrm{F}\) is \(V_{0}\) and the charge here is \(q_{0}\). We have
A) \(q_{0}=40 \mu C\)
B) \(V_{0}=60 \mathrm{~V}\)
C) \(V_{0}=80 \mathrm{~V}\)
D) \(q_{0}=240 \mu C\)

1 A, C are correct
2 A, B are correct
3 B, D are correct
4 C, D are correct
Electrostatic Potentials and Capacitance

268114 \(\mathbf{n}\) C apacitors of \(2 \mu \mathrm{F}\) each are connected in parallel and a p.d of \(200 \mathrm{v}\) is applied to the combination. The total charge on them was \(1 c\) then \(n\) is equal to

1 3333
2 3000
3 2500
4 25
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Electrostatic Potentials and Capacitance

268103 If \(\mathbf{3}\) capacitors of values 1,2 and \(3 \mu F\) are available. Themaximum and minimum values of capacitances one can obtain by different combinations of the three capacitors together are respectively .... and

1 \(6 \mu F, \frac{6}{11} \mu F\)
2 \(6 \mu F, \frac{11}{6} \mu F\)
3 \(3 \mu F, 1 \mu F\)
4 \(4 \mu F, 2 \mu F\)
Electrostatic Potentials and Capacitance

268111 Two condensers of capacity \(\mathrm{C}\) and \(2 \mathrm{C}\) are connected in parallel and these are charged upto \(\mathrm{V}\) volt. If the battery is removed and dielectric medium of constant \(K\) is put between the plates of first condenser, then the potential at each condenser is

1 \(\frac{V}{k+2}\)
2 \(2+\frac{k}{3 V}\)
3 \(\frac{2 V}{k+2}\)
4 \(\frac{3 V}{k+2}\)
Electrostatic Potentials and Capacitance

268112 Given a number of capacitors labelled as \(\mathrm{C}, \mathrm{V}\). Find the minimum number of capacitors needed to get an arrangement equivalent to \(C_{\text {net }}, V_{\text {net }}\)

1 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}{ }^{2}}{V^{2}}\)
2 \(n=\frac{C}{C_{\text {net }}} \times \frac{V^{2}}{V_{\text {net }}{ }^{2}}\)
3 \(n=\frac{C}{C_{n e t}} \times \frac{V}{V_{n e t}}\)
4 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}}{V}\)
Electrostatic Potentials and Capacitance

268113 Two capacitors of capacities \(3 \mu F\) and \(6 \mu F\) are connected in series and connected to 120V. The potential difference across \(3 \mu \mathrm{F}\) is \(V_{0}\) and the charge here is \(q_{0}\). We have
A) \(q_{0}=40 \mu C\)
B) \(V_{0}=60 \mathrm{~V}\)
C) \(V_{0}=80 \mathrm{~V}\)
D) \(q_{0}=240 \mu C\)

1 A, C are correct
2 A, B are correct
3 B, D are correct
4 C, D are correct
Electrostatic Potentials and Capacitance

268114 \(\mathbf{n}\) C apacitors of \(2 \mu \mathrm{F}\) each are connected in parallel and a p.d of \(200 \mathrm{v}\) is applied to the combination. The total charge on them was \(1 c\) then \(n\) is equal to

1 3333
2 3000
3 2500
4 25
Join With Us for More Updates
Electrostatic Potentials and Capacitance

268103 If \(\mathbf{3}\) capacitors of values 1,2 and \(3 \mu F\) are available. Themaximum and minimum values of capacitances one can obtain by different combinations of the three capacitors together are respectively .... and

1 \(6 \mu F, \frac{6}{11} \mu F\)
2 \(6 \mu F, \frac{11}{6} \mu F\)
3 \(3 \mu F, 1 \mu F\)
4 \(4 \mu F, 2 \mu F\)
Electrostatic Potentials and Capacitance

268111 Two condensers of capacity \(\mathrm{C}\) and \(2 \mathrm{C}\) are connected in parallel and these are charged upto \(\mathrm{V}\) volt. If the battery is removed and dielectric medium of constant \(K\) is put between the plates of first condenser, then the potential at each condenser is

1 \(\frac{V}{k+2}\)
2 \(2+\frac{k}{3 V}\)
3 \(\frac{2 V}{k+2}\)
4 \(\frac{3 V}{k+2}\)
Electrostatic Potentials and Capacitance

268112 Given a number of capacitors labelled as \(\mathrm{C}, \mathrm{V}\). Find the minimum number of capacitors needed to get an arrangement equivalent to \(C_{\text {net }}, V_{\text {net }}\)

1 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}{ }^{2}}{V^{2}}\)
2 \(n=\frac{C}{C_{\text {net }}} \times \frac{V^{2}}{V_{\text {net }}{ }^{2}}\)
3 \(n=\frac{C}{C_{n e t}} \times \frac{V}{V_{n e t}}\)
4 \(n=\frac{C_{n e t}}{C} \times \frac{V_{\text {net }}}{V}\)
Electrostatic Potentials and Capacitance

268113 Two capacitors of capacities \(3 \mu F\) and \(6 \mu F\) are connected in series and connected to 120V. The potential difference across \(3 \mu \mathrm{F}\) is \(V_{0}\) and the charge here is \(q_{0}\). We have
A) \(q_{0}=40 \mu C\)
B) \(V_{0}=60 \mathrm{~V}\)
C) \(V_{0}=80 \mathrm{~V}\)
D) \(q_{0}=240 \mu C\)

1 A, C are correct
2 A, B are correct
3 B, D are correct
4 C, D are correct
Electrostatic Potentials and Capacitance

268114 \(\mathbf{n}\) C apacitors of \(2 \mu \mathrm{F}\) each are connected in parallel and a p.d of \(200 \mathrm{v}\) is applied to the combination. The total charge on them was \(1 c\) then \(n\) is equal to

1 3333
2 3000
3 2500
4 25