Capacitance
Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $\mathrm{C}_{1} / \mathrm{C}_{2}$ is

1 $\frac{1}{4}$
2 $\frac{1}{16}$
3 $\frac{1}{8}$
4 $\frac{1}{12}$
Capacitance

165663 The effective capacitances of two capacitors are $3 \mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1 $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
2 $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
3 $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
4 $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$
Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1 $2 \mathrm{C}$
2 $\mathrm{C} / 2$
3 $4 \mathrm{C}$
4 $\mathrm{C} / 4$
Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathbf{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $\mathrm{C}_{0}=$ capacitance of capacitor with air as medium between plates.)

1 $\frac{4 \mathrm{KC}_{0}}{3+\mathrm{K}}$
2 $\frac{3 \mathrm{KC}_{0}}{3+\mathrm{K}}$
3 $\frac{3+\mathrm{K}}{4 \mathrm{KC}_{0}}$
4 $\frac{\mathrm{K}}{4+\mathrm{K}}$
Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $\mathrm{C}_{1} / \mathrm{C}_{2}$ is

1 $\frac{1}{4}$
2 $\frac{1}{16}$
3 $\frac{1}{8}$
4 $\frac{1}{12}$
Capacitance

165663 The effective capacitances of two capacitors are $3 \mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1 $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
2 $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
3 $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
4 $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$
Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1 $2 \mathrm{C}$
2 $\mathrm{C} / 2$
3 $4 \mathrm{C}$
4 $\mathrm{C} / 4$
Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathbf{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $\mathrm{C}_{0}=$ capacitance of capacitor with air as medium between plates.)

1 $\frac{4 \mathrm{KC}_{0}}{3+\mathrm{K}}$
2 $\frac{3 \mathrm{KC}_{0}}{3+\mathrm{K}}$
3 $\frac{3+\mathrm{K}}{4 \mathrm{KC}_{0}}$
4 $\frac{\mathrm{K}}{4+\mathrm{K}}$
Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $\mathrm{C}_{1} / \mathrm{C}_{2}$ is

1 $\frac{1}{4}$
2 $\frac{1}{16}$
3 $\frac{1}{8}$
4 $\frac{1}{12}$
Capacitance

165663 The effective capacitances of two capacitors are $3 \mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1 $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
2 $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
3 $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
4 $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$
Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1 $2 \mathrm{C}$
2 $\mathrm{C} / 2$
3 $4 \mathrm{C}$
4 $\mathrm{C} / 4$
Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathbf{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $\mathrm{C}_{0}=$ capacitance of capacitor with air as medium between plates.)

1 $\frac{4 \mathrm{KC}_{0}}{3+\mathrm{K}}$
2 $\frac{3 \mathrm{KC}_{0}}{3+\mathrm{K}}$
3 $\frac{3+\mathrm{K}}{4 \mathrm{KC}_{0}}$
4 $\frac{\mathrm{K}}{4+\mathrm{K}}$
Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $\mathrm{C}_{1} / \mathrm{C}_{2}$ is

1 $\frac{1}{4}$
2 $\frac{1}{16}$
3 $\frac{1}{8}$
4 $\frac{1}{12}$
Capacitance

165663 The effective capacitances of two capacitors are $3 \mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1 $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
2 $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
3 $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
4 $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$
Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1 $2 \mathrm{C}$
2 $\mathrm{C} / 2$
3 $4 \mathrm{C}$
4 $\mathrm{C} / 4$
Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathbf{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $\mathrm{C}_{0}=$ capacitance of capacitor with air as medium between plates.)

1 $\frac{4 \mathrm{KC}_{0}}{3+\mathrm{K}}$
2 $\frac{3 \mathrm{KC}_{0}}{3+\mathrm{K}}$
3 $\frac{3+\mathrm{K}}{4 \mathrm{KC}_{0}}$
4 $\frac{\mathrm{K}}{4+\mathrm{K}}$