In the given circuit reading of ideal voltmeter is $\frac{E}{2}$ then internal resistance of cell is ............. $\Omega$
Medium
Download our app for free and get started
$\frac{\varepsilon}{2}=\varepsilon-i r$ or $i=\frac{\varepsilon}{2 r}$
$2 \varepsilon=i(3+r)$
$r=1 \Omega$
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
- 1In the circuit shown in figure, all cells are ideal. The current through $2 \,\Omega$ resistor is ............ $A$View Solution
- 2In the circuit shown, what is the potential difference $V_{PQ}$? ................ $V$View Solution
- 3In the given potentiometer circuit, the resistance of the potentiometer wire $AB$ is $R_0$. $C$ is a cell of internal resistance $r$. The galvanometer $G$ does not give zero deflection for any position of the jockey $J$. Which of the following cannot be a reason for this?View Solution
- 4The current from the cell for the given circuit is ................ $\mathrm{A}$View Solution
- 5View SolutionWatt-hour meter measures
- 6A group of $N$ cells whose $emf$ varies directly with the internal resistance as per the equation $E_N = 1.5\, r_N$ are connected as shown in the figure below. The current $I$ in the circuit is ........... $amp$View Solution
- 7Two resistances are connected in two gaps of a meter bridge. The balance point is $20\, cm$ from the zero end. A resistance of $15\, ohms$ is connected in series with the smaller of the two. The null point shifts to $40\, cm$. The value of the smaller resistance in $ohms$ isView Solution
- 8A $d.c.$ main supply of $e.m.f.\, 220\, V$ is connected across a storage battery of $e.m.f.\, 200\, V$ through a resistance of $1\,\Omega $. The battery terminals are connected to an external resistance $'R'$ . The minimum value of $'R'$, so that a current passes through the battery to charge it is ............... $\Omega$View Solution
- 9In the circuit shown, $n$-identical resistors $R$ are connected in parallel $(n > 1)$ and the combination is connected in series to another resistor $R_0$. In the adjoining circuit $n$ resistors of resistance $R$ are all connected in series alongwith $R_0$. The batteries in both circuits are identical and net power dissipated in the $n$ resistors in both circuits is same. The ratio $R_0 / R$ isView Solution
- 10View SolutionIn the circuit shown in the figure, the current through
