For a wire $\frac{R}{l}=\frac{1}{2}$ and length of wire is $l=5\, cm .$ If potential difference $1\, V$ is applied across it, current through wire will be: $( R =$ Resistance $)$ (in $A$)
A$40$
B$4$
C$25$
D$2.5$
AIIMS 2019, Medium
Download our app for free and get started
A$40$
a The resistance of the wire is,
$R=\frac{l}{2}$
$=\frac{5 \times 10^{-2}}{2} \Omega$
The current through the wire will be,
$i=\frac{V}{R}$
$=\frac{1}{5 \times 10^{-2}} \times 2$
$=40 A$
Download our app
and 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.*
Figure shows a part of an electric circuit. The potentials at points $a , b$ and $c$ are $30\,V , 12\,V$ and $2\,V$ respectively. The current through the $20 \Omega$ resistor will be $........\,A$
In 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?
Two resistors are connected $(a)$ in series $(b)$ in parallel. The equivalent resistance in the two cases are $9$ $ohm$ and $2$ $ohm$ respectively. Then the resistances of the component resistors are
Potentiometer wire of length $1 \,m$ is connected in series with $490\,\Omega $ resistance and $2\,V$ battery. If $0.2\, mV/cm $ is the potential gradient, then resistance of the potentiometer wire is ................ $\Omega$