A wire when connected to $220\,V$ mains supply has power dissipation ${P_1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is ${P_2}$. Then ${P_2}:{P_1}$ is
A$1$
B$4$
C$2$
D$3$
AIEEE 2002, Medium
Download our app for free and get started
B$4$
b When wire is cut into two equal parts then power dissipated by each part is $2{P_1}$
So their parallel combination will dissipate power ${P_2} = 2{P_1} + 2{P_1} = 4{P_1}$
Which gives $\frac{{{P_2}}}{{{P_1}}} = 4$
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 wire $100\,cm$ long and $2.0\,mm$ diameter has a resistance of $0.7\, ohm$, the electrical resistivity of the material is ...........$ \times {10^{ - 6}}\,ohm \times m$
In the shown arrangement of the experiment of the meter bridge if $AC$ corresponding to null deflection of galvanometer is $x$, what would be its value if the radius of the wire $AB$ is doubled
Two cells are connected between points $A$ and $B$ as shown. Cell $1$ has emf of $12\,V$ and internal resistance of $3\,\Omega$. Cell $2$ has emf of $6\,V$ and internal resistance of $6\,\Omega$. An external resistor $R$ of $4\,\Omega$ is connected across $A$ and $B$. The current flowing through $R$ will be $.............A$.
An insulating pipe of cross-section area $'A'$ contains an electrolyte which has two types of ions $\rightarrow$ their charges being $-e$ and $+2e$. $A$ potential difference applied between the ends of the pipe result in the drifting of the two types of ions, having drift speed $= v$ ($-ve$ ion) and $v/4$ ($+ve$ ion). Both ions have the same number per unit volume $= n$. The current flowing through the pipe is
Two cells, having the same $e.m.f.$ are connected in series through an external resistance $R.$ Cells have internal resistances $r_1$ and $r_2\,\, (r_1 > r_2)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is
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......