Three capacitors of capacitances \(3~\mu\text{F}\), \(9~\mu\text{F}\) and \(18~\mu\text{F}\) are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases \(\frac{C_s}{C_p}\) will be:
1. \(1:15\)
2. \(15:1\)
3. \(1:1\)
4. \(1:3\)

Subtopic:  Combination of Capacitors |
 87%
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Four condensers each of capacity 4 μF are connected as shown in figure and VPVQ = 15 volts. The energy stored in the system is 

(1) 2400 ergs

(2) 1800 ergs

(3) 3600 ergs

(4) 5400 ergs

Subtopic:  Energy stored in Capacitor |
 83%
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In an adjoining figure are shown three capacitors C1, C2 and C3 joined to a battery. The correct condition will be (Symbols have their usual meanings) :

(1) Q1 = Q2 = Q3 and V1 = V2 = V3 = V

(2) Q1 = Q2 + Q3 and V = V1 + V2 + V3

(3) Q1 = Q2 + Q3 and V = V1 + V2

(4) Q2 = Q3 and V2 = V3

Subtopic:  Capacitance |
 67%
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In the circuit diagram shown in the adjoining figure, the resultant capacitance between P and Q is 

(1) 47 μF

(2) 3 μF

(3) 60 μF

(4) 10 μF

Subtopic:  Combination of Capacitors |
 89%
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Two capacitances of capacity C1 and C2 are connected in series and potential difference V is applied across it. Then the potential difference across C1 will be:

(1) VC2C1

(2) VC1+C2C1

(3) VC2C1+C2

(4) VC1C1+C2

Subtopic:  Combination of Capacitors |
 76%
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A capacitor of capacity C1 is charged to the potential of V0. On disconnecting with the battery, it is connected with a capacitor of capacity C2 as shown in the adjoining figure. The ratio of energies before and after the connection of switch S will be

(1) (C1 + C2)/C1

(2) C1/(C1 + C2)

(3) C1C2

(4) C1/C2

Subtopic:  Energy stored in Capacitor |
 56%
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Four capacitors each of capacity \(3~\mu\text{F}\) are connected as shown in the adjoining figure. The ratio of equivalent capacitance between \(A\) and \(B\) and between \(A\) and \(C\) will be:

       

1. \(4:3\)

2. \(3:4\)

3. \(2:3\)

4. \(3:2\)

Subtopic:  Combination of Capacitors |
 69%
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A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is \(A\) metre2 and the separation is \(t\) metre. The dielectric constants are \(k_1\) and \(k_2\) respectively. Its capacitance in farad will be:

        
1. \(\frac{\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
2. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} + k_{2}\right)}{2}\)
3. \(\frac{2\varepsilon_{0} A}{t} \left( k_{1} + k_{2}\right)\)
4. \(\frac{\varepsilon_{0} A}{t} \frac{\left( k_{1} - k_{2}\right)}{2}\)

Subtopic:  Dielectrics in Capacitors |
 62%
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Three condensers each of capacitance 2F are put in series. The resultant capacitance is 

(1) 6F

(2) 32F

(3) 23F

(4) 5F

Subtopic:  Combination of Capacitors |
 89%
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Four condensers are joined as shown in the adjoining figure. The capacity of each is 8 μF. The equivalent capacity between the points A and B will be

(1) 32 μF

(2) 2 μF

(3) 8 μF

(4) 16 μF

Subtopic:  Combination of Capacitors |
 75%
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