Two cells of emf $2\, E$ and $E$ with internal resistance $r _{1}$ and $r _{2}$ respectively are connected in series to an external resistor $R$ (see $figure$). The value of $R ,$ at which the potential difference across the terminals of the first cell becomes zero is
A$r _{1}+ r _{2}$
B$\frac{ r _{1}}{2}- r _{2}$
C$\frac{ r _{1}}{2}+ r _{2}$
D$r _{1}- r _{2}$
JEE MAIN 2021, Diffcult
Download our app for free and get started
B$\frac{ r _{1}}{2}- r _{2}$
b $i =\frac{3 E }{ R + r _{1}+ r _{2}}$
$TPD =2 E - ir _{1}=0$
$2 E = ir _{1}$
$2 E =\frac{3 E \times r _{1}}{ R + r _{1}+ r _{2}}$
$2 R +2 r _{1}+2 r _{2}=3 r _{1}$
$R =\frac{ r _{1}}{2}- r _{2}$
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.*
A copper wire of length $10\,m$ and radius $\left(10^{-2} / \sqrt{\pi}\right) m$ has electrical resistance of $10 \,\Omega$. The current density in the wire for an electric field strength of $10( V / m )$ is :
When a resistor of $11 \,\Omega$ is connected in series with an electric cell, the current flowing in it is $0.5\, A$. Instead, when a resistor of $5 \,\Omega$ is connected to the same electric cell in series, the current increases by $0.4\, A$. The internal resistance of the cell is ................ $\Omega$