Four resistances $40 \ \Omega, 60\ \Omega, 90\ \Omega$ and $110\ \Omega$ make the arms of a quadrilateral $A,B,C,D$. Across $AC$ is a battery of emf $40\, V$ and internal resistance negligible. The potential difference across $BD$ is $V$ is......
A$4$
B$1$
C$2$
D$5$
JEE MAIN 2020, Medium
Download our app for free and get started
C$2$
c $i _{1}=\frac{40}{40+60}=0.4$
$i _{2}=\frac{40}{90+110}=\frac{1}{5}$
$v _{ B }+ i _{1}(40)- i _{2}(90)= v _{ D }$
$v _{ B }- v _{ D }=\frac{1}{5}(90)-\frac{4}{10} \times 40$
$v _{ B }- v _{ D }=18-16=2$
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.*
Ametallic conductor of irregular cross-section is as shown in the figure. Aconstant potential difference is applied across the ends $(1)$ and $(2)$. Then :
If in the circuit shown below, the internal resistance of the battery is $1.5$ $\Omega$ and $V_P$ and $V_Q$ are the potentials at $P$ and $Q$ respectively, what is the potential difference between the points $P$ and $Q$
Five identical cells each of internal resistance $1\, \Omega$ and $emf \;5\, {V}$ are connected in series and in parallel with an external resistance $'R'.$ For what value of $'R',$ current in series and parallel combination will remain the same ? (in $\Omega$)
Consider a block of conducting material ofresistivity '$\rho$' shown in the figure. Current '$I$' enters at '$A$' and leaves from '$D$'. We apply superp osition principle to find voltage '$\Delta V$ ' developed between '$B$' and '$C$'. The calculation is done in the following steps:
$(i)$ Take current '$I$' entering from '$A$' and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculatefield $E(r)$ at distance '$r$' from $A$ by using Ohm's law $E = \rho j$, where j is the current per unit area at '$r$'.
(iii) From the '$r$' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat $(i), (ii)$ and $(iii)$ for current '$I$' leaving '$D$' and superpose results for '$A$' and '$D$'.
For what value of unknown resistance $X$, the potential difference between $B$ and $D$ will be zero in the circuit shown in the figure ............... $\Omega$
The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}({E_1} > {E_2})$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$, the null point will be