A conducting wire of length $ l$ area of cross-section $A$ and electric resistivity $\rho$ is connected between the terminals of a battery. $A$ potential difference $V$ is developed between its ends, causing an electric current.If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be
JEE MAIN 2021, Medium
Download our app for free and get started
As per the question
Resistance $=\frac{\rho(2 l)}{( A / 2)}=\frac{4 \rho l}{ A }$
$\Rightarrow$ Current $=\frac{ V }{ R }=\frac{ VA }{4 \rho l}$
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.*
When the current $i$ is flowing through a conductor, the drift velocity is $v$. If $2i$ current is flowed through the same metal but having double the area of cross-section, then the drift velocity will be
A cell of internal resistance $r$ drives current through an external resistance $R$ . The power delivered by the cell to the external resistance will be maximum when:
In the circuit shown below, the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q _1 \mu C$. Then $S$ is switched to position $Q$. After a long time, the charge on the capacitor is $q _2 \mu C$.
Current $l$ versus time $t$ graph through a conductor is shown in the figure. Average current through the conductor in the interval $0$ to $15 \,s$ is ............ $A$
If an electron revolves in the path of a circle of radius of $0.5 × 10^{-10}\, m$ at frequency of $5 × 10^{15}$ $cycles/s$ the electric current in the circle is ..................$mA$ (Charge of an electron $=1.6 × 10^{-19}\, C$ )
When two resistance $R_1$ and $R_2$ connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \Omega$ is introduced into the right gap, a null point is found at $60 cm$ from left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap, a resistance of $3 \Omega$ is introduced into the right-gap to get null point at 40 cm from left end. The product of $R_1 R_2$ is $.............\Omega$