A cell of emf $6\, V$ and resistance $0.5\, ohm$ is short circuited. The current in the cell is ............... $amp$
Easy
Download our app for free and get started
(b) $i = \frac{E}{r} = \frac{6}{{0.5}} = 12\,amp.$
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 an electrolyte $3.2 \times {10^{18}}$ bivalent positive ions drift to the right per second while $3.6 \times {10^{18}}$ monovalent negative ions drift to the left per second. Then the current isView Solution
- 2View SolutionA potential difference is applied across the ends of a metallic wire. If the potential difference is doubled, then the drift velocity
- 3$ABCD$ is a square where each side is a uniform wire of resistance $1\,\Omega$ . $A$ point $E$ lies on $CD$ such that if a uniform wire of resistance $1\,\Omega$ is connected across $AE$ and constant potential difference is applied across $A$ and $C$ then $B$ and $E$ are equipotential.View Solution
- 4There is a current of $20$ amperes in a copper wire of ${10^{ - 6}}$ square meter area of cross-section. If the number of free electrons per cubic meter is ${10^{29}}$, then the drift velocity isView Solution
- 5A steady current $I$ flows through a wire of radius $r$, length $L$ and resistivity $\rho$. The current produces heat in the wire. The rate of heat loss in a wire is proportional to its surface area. The steady temperature of the wire is independent ofView Solution
- 6The resistance of the meter bridge $AB$ in given figure is $4\,\Omega $. With a cell of emf $\varepsilon \, = 0.5\,\,V$ and rheostat resistance $R_h = 2\,\Omega $ the null point is obtained at some point $J.$ When the cell is replaced by another one of emf $\varepsilon \, = {\varepsilon _2}$ the same null point $J$ is found for $R_h = 6\,\Omega .$ The $emf$ ${\varepsilon _2}$ is ................. $V$View Solution
- 7Drift speed of electrons, when $1.5\, A$ of current flows in a copper wire of cross section $5\, mm^2$, is $v$. If the electron density in copper is $9 \times 10^{28}\, m^3$ the value of $v$ in $mm/s$ is close to (Take charge of electron to be $= 1.6 \times 10^{-19}\, C$)View Solution
- 8In the circuit here, the steady state voltage across capacitor $C$ is a fraction of the battery $e.m.f.$ The fraction is decided byView Solution
- 9View SolutionIn the circuit shown in the figure, the current through
- 10View SolutionIn the given circuit, with steady current, the potential drop across the capacitor must be
