In the given circuit, with steady current, the potential drop across the capacitor must be
Medium
Download our app for free and get started
in sleady state
$I=\frac{2 V-V}{2 R+R}=\frac{V}{3 R}$
Potential differnce across $"\mathrm{R}"$ $=\frac{\mathrm{V}}{3 \mathrm{R}} \times \mathrm{R}=\frac{\mathrm{V}}{3}$
$\therefore \mathrm{V}_{\mathrm{AB}}=\mathrm{V}+\frac{\mathrm{V}}{3}$
and
So potential difference across $\mathrm{C}=\frac{\mathrm{V}}{3}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A battery of $6\,V$ is connected to the circuit as shown below. The current I drawn from the battery is:View Solution
- 2Two circuits (shown below) are called ‘Circuit $A$ ’and ‘Circuit $B$’. The equivalent resistance of ‘Circuit $a$’ is $x$ and that of ‘Circuit $B$’ is $y$ between $1$ and $2.$View Solution
- 3A potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given byView Solution
- 4In the figure shown the power generated in $y$ is maximum when $y = 5\,\Omega $. Then $R$ isView Solution
- 5When the resistance $R$ (indicated in the figure below) is changed from $1 \,k \Omega$. to $10 \,k \Omega$, the current flowing through the resistance $R'$ does not change. What is the value of the resistor $R'?$View Solution
- 6The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\,\Omega $ , the deflection in galvanometer becomes $\frac{{\theta _0}}{5}$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: .................. $\Omega$View Solution
- 7Find current through the branch $ad$ of the circuit shown in figure. :-............. $A$View Solution
- 8In the circuit shown, the cell has $emf = 10\,V$ and internal resistance $= 1\, \Omega$ :-View Solution
- 9View SolutionElectromotive force is the force, which is able to maintain a constant
- 10In the network shown in the following figure, each resistance is $1\,ohm.$ The effective resistance between $A$ and $B$ isView Solution
