Effect of Dielectric Charging and Discharging of Capacitor
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Capacitance

166007 Calculate charge on capacitor in steady state.

1 $50 \mu \mathrm{C}$
2 $30 \mu \mathrm{C}$
3 $45 \mu \mathrm{C}$
4 $60 \mu \mathrm{C}$
Capacitance

166008 What is the energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ ?

1 $0.4 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $0.05 \mathrm{~J}$
4 $0.1 \mathrm{~J}$
Capacitance

166009 In the adjoining figure, $E=5 \mathrm{~V}, \mathrm{r}=1 \Omega, \mathrm{R}_{2}=$ $4 \Omega, R_{1}=R_{3}=1 \Omega$ and $C=3 \mu F$. The numerical value of the charge on each plate of the capacitor is

1 $3 \mu \mathrm{C}$
2 $6 \mu \mathrm{C}$
3 $12 \mu \mathrm{C}$
4 $24 \mu \mathrm{C}$
Capacitance

166010 A voltage $V_{P Q}=V_{0} \cos \omega t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c=\frac{1}{\omega R \sqrt{3}} \text { and } L=\frac{R \sqrt{3}}{\omega}$
Then, the total impedance between $P$ and $Q$ is

1 $1.5 \mathrm{R}$
2 $2 \mathrm{R}$
3 $3 \mathrm{R}$
4 $4 \mathrm{R}$
5 $2.5 \mathrm{R}$
Capacitance

166007 Calculate charge on capacitor in steady state.

1 $50 \mu \mathrm{C}$
2 $30 \mu \mathrm{C}$
3 $45 \mu \mathrm{C}$
4 $60 \mu \mathrm{C}$
Capacitance

166008 What is the energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ ?

1 $0.4 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $0.05 \mathrm{~J}$
4 $0.1 \mathrm{~J}$
Capacitance

166009 In the adjoining figure, $E=5 \mathrm{~V}, \mathrm{r}=1 \Omega, \mathrm{R}_{2}=$ $4 \Omega, R_{1}=R_{3}=1 \Omega$ and $C=3 \mu F$. The numerical value of the charge on each plate of the capacitor is

1 $3 \mu \mathrm{C}$
2 $6 \mu \mathrm{C}$
3 $12 \mu \mathrm{C}$
4 $24 \mu \mathrm{C}$
Capacitance

166010 A voltage $V_{P Q}=V_{0} \cos \omega t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c=\frac{1}{\omega R \sqrt{3}} \text { and } L=\frac{R \sqrt{3}}{\omega}$
Then, the total impedance between $P$ and $Q$ is

1 $1.5 \mathrm{R}$
2 $2 \mathrm{R}$
3 $3 \mathrm{R}$
4 $4 \mathrm{R}$
5 $2.5 \mathrm{R}$
Capacitance

166007 Calculate charge on capacitor in steady state.

1 $50 \mu \mathrm{C}$
2 $30 \mu \mathrm{C}$
3 $45 \mu \mathrm{C}$
4 $60 \mu \mathrm{C}$
Capacitance

166008 What is the energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ ?

1 $0.4 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $0.05 \mathrm{~J}$
4 $0.1 \mathrm{~J}$
Capacitance

166009 In the adjoining figure, $E=5 \mathrm{~V}, \mathrm{r}=1 \Omega, \mathrm{R}_{2}=$ $4 \Omega, R_{1}=R_{3}=1 \Omega$ and $C=3 \mu F$. The numerical value of the charge on each plate of the capacitor is

1 $3 \mu \mathrm{C}$
2 $6 \mu \mathrm{C}$
3 $12 \mu \mathrm{C}$
4 $24 \mu \mathrm{C}$
Capacitance

166010 A voltage $V_{P Q}=V_{0} \cos \omega t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c=\frac{1}{\omega R \sqrt{3}} \text { and } L=\frac{R \sqrt{3}}{\omega}$
Then, the total impedance between $P$ and $Q$ is

1 $1.5 \mathrm{R}$
2 $2 \mathrm{R}$
3 $3 \mathrm{R}$
4 $4 \mathrm{R}$
5 $2.5 \mathrm{R}$
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Capacitance

166007 Calculate charge on capacitor in steady state.

1 $50 \mu \mathrm{C}$
2 $30 \mu \mathrm{C}$
3 $45 \mu \mathrm{C}$
4 $60 \mu \mathrm{C}$
Capacitance

166008 What is the energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ ?

1 $0.4 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $0.05 \mathrm{~J}$
4 $0.1 \mathrm{~J}$
Capacitance

166009 In the adjoining figure, $E=5 \mathrm{~V}, \mathrm{r}=1 \Omega, \mathrm{R}_{2}=$ $4 \Omega, R_{1}=R_{3}=1 \Omega$ and $C=3 \mu F$. The numerical value of the charge on each plate of the capacitor is

1 $3 \mu \mathrm{C}$
2 $6 \mu \mathrm{C}$
3 $12 \mu \mathrm{C}$
4 $24 \mu \mathrm{C}$
Capacitance

166010 A voltage $V_{P Q}=V_{0} \cos \omega t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c=\frac{1}{\omega R \sqrt{3}} \text { and } L=\frac{R \sqrt{3}}{\omega}$
Then, the total impedance between $P$ and $Q$ is

1 $1.5 \mathrm{R}$
2 $2 \mathrm{R}$
3 $3 \mathrm{R}$
4 $4 \mathrm{R}$
5 $2.5 \mathrm{R}$