Q. No.From the circuit given below, the capacitance between terminals \(A\) and \(B\) shown in the circuit is: (in \(\mu\text{F}\))
(take \(C_1=C_2=C_3=1~\mu\text{F}\) and \(C_4 = 2~\mu\text{F}\))
1. \(2\)
2. \(7/2\)
3. \(7/3\)
4. \(5/2\)
(take \(C_1=C_2=C_3=1~\mu\text{F}\) and \(C_4 = 2~\mu\text{F}\))
1. \(2\)
2. \(7/2\)
3. \(7/3\)
4. \(5/2\)
Subtopic: Combination of Capacitors |
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Four capacitors each of capacitance \(16 ~\mu \text{F}\) are connected as shown in the figure. The capacitance between points \(A\) and \(B\) is: (in \(\mu \text{F}\)).
1. \(64\)
2. \(50\)
3. \(44\)
4. \(30\)
1. \(64\)
2. \(50\)
3. \(44\)
4. \(30\)
Subtopic: Combination of Capacitors |
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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 \(A/3\) and the height is d, the capacitance of the arrangement is :
1. \(\frac{18 \epsilon_{\mathrm{o}} \mathrm{~A}}{11 \mathrm{~d}}\)
2. \(\frac{11 \epsilon_0 A}{20 d}\)
3. \(\frac{13 \epsilon_0 A}{17 d}\)
4. \(\frac{11 \epsilon_0 A}{18 d}\)
1. \(\frac{18 \epsilon_{\mathrm{o}} \mathrm{~A}}{11 \mathrm{~d}}\)
2. \(\frac{11 \epsilon_0 A}{20 d}\)
3. \(\frac{13 \epsilon_0 A}{17 d}\)
4. \(\frac{11 \epsilon_0 A}{18 d}\)
Subtopic: Combination of Capacitors |
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What will be the equivalent capacitance across points \(A\) and \(B\) in the given electrical circuit?
| 1. | \(\dfrac{C}{2}\) | 2. | \(2C\) |
| 3. | \(\dfrac{5C}{3}\) | 4. | \(\dfrac{3C}{4}\) |
Subtopic: Combination of Capacitors |
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A parallel plate capacitor is made up of a stair-like structure with a plate area \(A\) of each stair and that is connected with a wire of length \(b\), as shown in the figure. The capacitance of the arrangement is \(\frac{x}{15} \frac{\varepsilon_0 A}{b}\). The value of \(x \) is:
1. \(13\)
2. \(23\)
3. \(33\)
4. \(43\)
1. \(13\)
2. \(23\)
3. \(33\)
4. \(43\)
Subtopic: Combination of Capacitors |
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The charge on the capacitor of capacitance \(15~ \mu \text F\) in the figure given below is:
| 1. | \(60~ \mu\text C\) | 2. | \(130 ~\mu\text C\) |
| 3. | \(260~ \mu \text C\) | 4. | \(585 ~\mu \text C\) |
Subtopic: Combination of Capacitors |
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Two capacitors having capacitance \(C_1\) and \(C_2\) respectively are connected as shown in the figure. Initially, capacitor \(C_1\) is charged to a potential difference \(V\) volt by a battery. The battery is then removed and the charged capacitor \(C_1\) is now connected to uncharged capacitor \(C_2\) by closing the switch \(S\). The amount of charge on the capacitor \(C_2\), after equilibrium, is:
1. \(\dfrac{C_1 C_2}{\left(C_1+C_2\right)} V \)
2. \(\dfrac{\left(C_1+C_2\right)}{C_1 C_2} V \)
3. \(\left({C}_1+{C}_2\right) {V} \)
4. \(\left({C}_1-{C}_2\right) {V}\)
1. \(\dfrac{C_1 C_2}{\left(C_1+C_2\right)} V \)
2. \(\dfrac{\left(C_1+C_2\right)}{C_1 C_2} V \)
3. \(\left({C}_1+{C}_2\right) {V} \)
4. \(\left({C}_1-{C}_2\right) {V}\)
Subtopic: Combination of Capacitors |
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Two metallic plates form a parallel plate capacitor. The distance between the plates is \(d\). A metal sheet of thickness \( d \over 2\) and area equal to the area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?
1. \(2:1\)
2. \(1:2\)
3. \(1:4\)
4. \(4:1\)
1. \(2:1\)
2. \(1:2\)
3. \(1:4\)
4. \(4:1\)
Subtopic: Combination of Capacitors |
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The equivalent capacitance between points A and B in the figure (shown below) will be:
1. \(2~\mu \text F\)
2. \(4~\mu \text F\)
3. \(6~\mu \text F\)
4. \(8~\mu \text F\)
1. \(2~\mu \text F\)
2. \(4~\mu \text F\)
3. \(6~\mu \text F\)
4. \(8~\mu \text F\)
Subtopic: Combination of Capacitors |
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A composite parallel plate capacitor is made up of two different dielectric materials with different thicknesses ( \(t_1\) and \(t_2\)) as shown in the figure. The two different dielectric materials are separated by a conducting foil \(F\). The voltage of the conducting foil is:
1. \(20\) V
2. \(40\) V
3. \(80\) V
4. \(60\) V
1. \(20\) V
2. \(40\) V
3. \(80\) V
4. \(60\) V
Subtopic: Combination of Capacitors |
Level 3: 35%-60%
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