In the given network capacitance, C1 = 10 μF, C2 = 5 μF and C3 = 4 μF. What is the resultant capacitance between A and B
1. 2.2 μF
2. 3.2 μF
3. 1.2 μF
4. 4.7 μF
The equivalent capacitance between AA and BB is:
1. | 2 μF2 μF | 2. | 3 μF3 μF |
3. | 5 μF5 μF | 4. | 0.5 μF0.5 μF |
In the circuit shown in figure, each capacitor has a capacity of 3 μF. The equivalent capacity between A and B is
1. 34μF34μF
2. 3 μF
3. 6 μF
4. 5 μF
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. 2ε0Ad [1k1+1k2+1k3]2ε0Ad [1k1+1k2+1k3]
2. ε0Ad [1k1+1k2+1k3]ε0Ad [1k1+1k2+1k3]
3. 2ε0Ad [k1+k2+k3]2ε0Ad [k1+k2+k3]
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 C1=2 μFC1=2 μF and C2=6 μFC2=6 μF in series, are connected in parallel to a third capacitor C3=4 μFC3=4 μF. This arrangement is then connected to a battery of emf=2 Vemf=2 V, as shown in the figure. How much energy is lost by the battery in charging the capacitors?
1. 22×10−6 J22×10−6 J
2. 11×10−6 J11×10−6 J
3. 323×10−6 J323×10−6 J
4. 163×10−6 J163×10−6 J
A parallel plate capacitor has capacitance CC. If it is equally filled with parallel layers of materials of dielectric constants K1K1 and K2K2, its capacity becomes C1C1. The ratio of C1C1 to CC is:
1. | K1+K2K1+K2 | 2. | K1K2K1−K2K1K2K1−K2 |
3. | K1+K2K1K2K1+K2K1K2 | 4. | 2K1K2K1+K22K1K2K1+K2 |
The equivalent capacitance in the circuit between A and B will be
1. 1 μF
2. 2 μF
3. 3 μF
4. 13 μF13 μF