There are a large number of cells available, each marked $(6 \,V , 0.5 \,\Omega)$ to be used to supply current to a device of resistance $0.75 \,\Omega$, requiring $24 \,A$ current. How should the cells be arranged, so that power is transmitted to the load using minimum number of cells?
Medium
Download our app for free and get started
(b)
$E=6 \,V$
$r=0.5 \,\Omega$
$R=0.75 \,\Omega$
$i=24 \,\Omega$
$S(0.5)=P(0.75)$
$2 \,s=3 p$
$i=\frac{P S E}{S r+P R}$
$24=\frac{P\left(\frac{3}{2} P\right) 6}{15 P}$
$P=4$ rows
$S=6$ cells
Download our appand 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.*
Similar Questions
- 1If two electric bulbs have $40\,W$ and $60\,W$ rating at $220\,V$, then the ratio of their resistances will beView Solution
- 2View SolutionThe example for non-ohmic resistance is
- 3The equivalent resistance of the circuit shown in the figure is ............. $\Omega$View Solution
- 4$12$ cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells identical with each other and also identical with the previous cells. The current is $3\, A$ when the external cells aid this battery and is $2\,A$ when the cells oppose the battery. How many cells in the battery are wrongly connected :-View Solution
- 5Four wires $AB,\,\,BC,\,\,CD,\,\,DA$ of resistance $4\, \Omega$ each and a fifth wire $BD$ of resistance $8\, \Omega$ are joined to form a rectangle $ABCD$ of which $BD$ is a diagonal. The effective resistance between the points $A$ and $B$ isView Solution
- 6Four ammeters with identical internal resistances $r$ and a resistor of resistance $R$ are connected to a current source as shown in figure. It is known that the reading of the ammeter $A_1$ is $I_1 = 3\ A$ and the reading of the ammeter $A_2$ is $I_2 = 5\ A$ . Determine the ratio of the resistances $R/r$ .View Solution
- 7Assertion : The current density $\vec J$ at any point in ohmic resistor is in direction of electric field $\vec E$ at that point.View Solution
Reason : A point charge when released from rest in a region having only electrostatic field always moves along electric lines of force.
- 8The length of a potentiometer wire is $1200\; \mathrm{cm}$ and it carries a current of $60 \;\mathrm{mA}$. For a cell of $emf\;5\; \mathrm{V}$ and intemal resistance of $20\; \Omega,$ the null point on it is found to be a $1000\; \mathrm{cm} .$ The resistance of whole wire is .............. $\Omega$View Solution
- 9View SolutionSelect the correct statement
- 10View SolutionAssertion : Kirchoff 's junction rule follows from conservation of charge.
Reason : Kirchoff 's loop rule follows from conservation of momentum.