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Three capacitors each of \({4}~\mu \text{F}\) are to be connected in such a way that the effective capacitance is \({6}~\mu \text{F}.\) This can be done by connecting them:
1. connecting all of them in a series
2. connecting all of them in parallel
3. connecting two in series and one in parallel
4. connecting two in parallel and one in series
Q1:
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Subtopic:  Combination of Capacitors |
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The equivalent capacitance between \(A\) and \(B \) in the circuit given below is:

1. \(3.6~\mu \text F\)
2. \(2.4~\mu \text F\)
3. \(4.9~\mu \text F\)
4. \(5.4~\mu \text F\)
Q2:
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Figure shows charge (\(q\)) versus voltage (\(V\)) graph for series and parallel combination of two given capacitors. The capacitance are:

                  
1. \(50~ \mu \text{F} \text { and } 30 ~\mu \text{F}\)
2. \(40~ \mu \text{F} \text { and } 10 ~\mu \text{F}\)
3. \(20~ \mu \text{F} \text { and } 30 ~\mu \text{F}\)
4. \(60~ \mu \text{F} \text { and } 40 ~\mu \text{F}\)

Q3:
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Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacitances, \(\left ( \dfrac{C_{Series}}{C_{Parallel}} \right )\) in these two cases will be:
1. \(4:1\)
2. \(2:1\)
3. \(1:4\)
4. \(1:2\)

Q4:
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Consider the combination of 2 capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\), with \(\mathrm{C}_2>\mathrm{C}_1 \), when connected in parallel, the equivalent capacitance is \(\frac{15}{4}\)times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors,\(\frac{\mathrm{C}_2}{\mathrm{C}_1}\).
1. \(\frac{15}{11}\)
2. \(\frac{111}{80}\)
3. \(\frac{29}{15}\)
4. None of the above

Q5:
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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. \(8~\mu \text F\)
Q6:
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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. \(4:1\)
Q7:
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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. \(\left({C}_1-{C}_2\right) {V}\)
Q8:
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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. \(585 ~\mu \text C\)
Q9:
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A parallel plate capacitor with plate area \(A\) and plate separation \(d =2~\text{m}\) has a capacitance of \( 4~\mu\text{F}\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K=3\) (as shown in the figure) will be:
               
1. \( 2~\mu\text{F}\)
2. \( 32~\mu\text{F}\)
3. \( 6~\mu\text{F}\)
4. \( 8~\mu\text{F}\)
Q10:
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