165791 The equivalent capacitance of the combination shown is

165792 The four capacitors, each of $25 \mu \mathrm{F}$ are connected as shown in fig. The dc voltmeter reads $200 \mathrm{~V}$. The charge on each plate of Capacitor is

165793 A network of four capacitors of capacity equal to $\mathrm{C}_{1}=\mathrm{C}, \mathrm{C}_{2}=2 \mathrm{C}, \mathrm{C}_{3}=3 \mathrm{C}$ and $\mathrm{C}_{4}=4 \mathrm{C}$ are conducted to a battery as shown in the figure . The ratio of the charges on $\mathrm{C}_{2}$ and $\mathrm{C}_{4}$ is

165794 Two capacitors each having a capacitance $2 \times 10^{-6} \mathrm{~F}$ and a breakdown voltage $5000 \mathrm{~V}$, are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

165795 A circuit is shown in the following figure for which $\mathbf{C}_{1}=(3 \pm 0.011) \mu \mathbf{F}, \mathbf{C}_{2}=(5 \pm 0.01) \mu \mathbf{F}$ and $\mathbf{C}_{3}=(1 \pm 0.01) \mu \mathbf{F}$. If $\mathbf{C}$ is the equivalent capacitance across $A B$, then $C$ is given by:

165791 The equivalent capacitance of the combination shown is

165792 The four capacitors, each of $25 \mu \mathrm{F}$ are connected as shown in fig. The dc voltmeter reads $200 \mathrm{~V}$. The charge on each plate of Capacitor is

165793 A network of four capacitors of capacity equal to $\mathrm{C}_{1}=\mathrm{C}, \mathrm{C}_{2}=2 \mathrm{C}, \mathrm{C}_{3}=3 \mathrm{C}$ and $\mathrm{C}_{4}=4 \mathrm{C}$ are conducted to a battery as shown in the figure . The ratio of the charges on $\mathrm{C}_{2}$ and $\mathrm{C}_{4}$ is

165794 Two capacitors each having a capacitance $2 \times 10^{-6} \mathrm{~F}$ and a breakdown voltage $5000 \mathrm{~V}$, are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

165795 A circuit is shown in the following figure for which $\mathbf{C}_{1}=(3 \pm 0.011) \mu \mathbf{F}, \mathbf{C}_{2}=(5 \pm 0.01) \mu \mathbf{F}$ and $\mathbf{C}_{3}=(1 \pm 0.01) \mu \mathbf{F}$. If $\mathbf{C}$ is the equivalent capacitance across $A B$, then $C$ is given by:

165791 The equivalent capacitance of the combination shown is

165792 The four capacitors, each of $25 \mu \mathrm{F}$ are connected as shown in fig. The dc voltmeter reads $200 \mathrm{~V}$. The charge on each plate of Capacitor is

165793 A network of four capacitors of capacity equal to $\mathrm{C}_{1}=\mathrm{C}, \mathrm{C}_{2}=2 \mathrm{C}, \mathrm{C}_{3}=3 \mathrm{C}$ and $\mathrm{C}_{4}=4 \mathrm{C}$ are conducted to a battery as shown in the figure . The ratio of the charges on $\mathrm{C}_{2}$ and $\mathrm{C}_{4}$ is

165794 Two capacitors each having a capacitance $2 \times 10^{-6} \mathrm{~F}$ and a breakdown voltage $5000 \mathrm{~V}$, are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

165795 A circuit is shown in the following figure for which $\mathbf{C}_{1}=(3 \pm 0.011) \mu \mathbf{F}, \mathbf{C}_{2}=(5 \pm 0.01) \mu \mathbf{F}$ and $\mathbf{C}_{3}=(1 \pm 0.01) \mu \mathbf{F}$. If $\mathbf{C}$ is the equivalent capacitance across $A B$, then $C$ is given by:

165791 The equivalent capacitance of the combination shown is

165792 The four capacitors, each of $25 \mu \mathrm{F}$ are connected as shown in fig. The dc voltmeter reads $200 \mathrm{~V}$. The charge on each plate of Capacitor is

165793 A network of four capacitors of capacity equal to $\mathrm{C}_{1}=\mathrm{C}, \mathrm{C}_{2}=2 \mathrm{C}, \mathrm{C}_{3}=3 \mathrm{C}$ and $\mathrm{C}_{4}=4 \mathrm{C}$ are conducted to a battery as shown in the figure . The ratio of the charges on $\mathrm{C}_{2}$ and $\mathrm{C}_{4}$ is

165794 Two capacitors each having a capacitance $2 \times 10^{-6} \mathrm{~F}$ and a breakdown voltage $5000 \mathrm{~V}$, are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

165795 A circuit is shown in the following figure for which $\mathbf{C}_{1}=(3 \pm 0.011) \mu \mathbf{F}, \mathbf{C}_{2}=(5 \pm 0.01) \mu \mathbf{F}$ and $\mathbf{C}_{3}=(1 \pm 0.01) \mu \mathbf{F}$. If $\mathbf{C}$ is the equivalent capacitance across $A B$, then $C$ is given by:

165791 The equivalent capacitance of the combination shown is

165792 The four capacitors, each of $25 \mu \mathrm{F}$ are connected as shown in fig. The dc voltmeter reads $200 \mathrm{~V}$. The charge on each plate of Capacitor is

165793 A network of four capacitors of capacity equal to $\mathrm{C}_{1}=\mathrm{C}, \mathrm{C}_{2}=2 \mathrm{C}, \mathrm{C}_{3}=3 \mathrm{C}$ and $\mathrm{C}_{4}=4 \mathrm{C}$ are conducted to a battery as shown in the figure . The ratio of the charges on $\mathrm{C}_{2}$ and $\mathrm{C}_{4}$ is

165794 Two capacitors each having a capacitance $2 \times 10^{-6} \mathrm{~F}$ and a breakdown voltage $5000 \mathrm{~V}$, are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

165795 A circuit is shown in the following figure for which $\mathbf{C}_{1}=(3 \pm 0.011) \mu \mathbf{F}, \mathbf{C}_{2}=(5 \pm 0.01) \mu \mathbf{F}$ and $\mathbf{C}_{3}=(1 \pm 0.01) \mu \mathbf{F}$. If $\mathbf{C}$ is the equivalent capacitance across $A B$, then $C$ is given by: