In the circuit shown in figure, each capacitor has a capacity of 3 μF. The equivalent capacity between A and B is
(1)
(2) 3 μF
(3) 6 μF
(4) 5 μF
Two capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then
1. The potential difference across the plates of A is 4V and across the plates of B is 6V
2. The potential difference across the plates of A is 6V and across the plates of B is 4V
3. The ratio of electrical energies stored in A and B is 2 : 3
4. The ratio of charges on A and B is 3 : 2
In the figure, three capacitors each of capacitance 6 pF are connected in series. The total capacitance of the combination will be
(1) 9 × 10–12 F
(2) 6 × 10–12 F
(3) 3 × 10–12 F
(4) 2 × 10–12 F
Equivalent capacitance between A and B is
(1) 8 μF
(2) 6 μF
(3) 26 μF
(4) 10/3 μF
In the figure a capacitor is filled with dielectrics. The resultant capacitance is
(1)
(2)
(3)
(4) None of these
Three capacitors of capacitance 3 μF, 10 μF and 15 μF are connected in series to a voltage source of 100V. The charge on 15 μF is
(1) 50 μC
(2) 100 μC
(3) 200 μC
(4.) 280 μC
Two capacitors \(C_1 = 2~\mu\text{F}\) and \(C_2 = 6~\mu \text{F}\) in series, are connected in parallel to a third capacitor \(C_3= 4~\mu\text{F}\). This arrangement is then connected to a battery of \(\text{emf}= 2~\text{V}\), as shown in the figure. How much energy is lost by the battery in charging the capacitors?
1. \(22\times 10^{-6}~\text{J}\)
2. \(11\times 10^{-6}~\text{J}\)
3. \(\frac{32}{3}\times 10^{-6}~\text{J}\)
4. \(\frac{16}{3}\times 10^{-6}~\text{J}\)
A parallel plate capacitor has capacitance \(C\). If it is equally filled with parallel layers of materials of dielectric constants \(K_1\) and \(K_2\), its capacity becomes \(C_1\). The ratio of \(C_1\) to \(C\) is:
1. | \(K_1 + K_2\) | 2. | \(\frac{K_{1} K_{2}}{K_{1}-K_{2}}\) |
3. | \(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\) | 4. | \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\) |
The equivalent capacitance in the circuit between A and B will be
(1) 1 μF
(2) 2 μF
(3) 3 μF
(4)
The equivalent capacitance between A and B is
(1)
(2)
(3)
(4)