In the circuit shown in the figure, the current through
IIT 1998, Medium
Download our app for free and get started
Equivalent resistance of the circuit $R = 9\,\Omega $
Main current $i = \frac{V}{R} = \frac{9}{9} = 1\,A$
After proper distribution, the current through $4\,\Omega$ resistance is $0.25\, A.$
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.*
In a building there are $15$ bulbs of $45\; \mathrm{W}, 15$ bulbs of $100\; \mathrm{W}, 15$ small fans of $10 \;\mathrm{W}$ and $2$ heaters of $1 \;\mathrm{kW}$. The voltage of electric main is $220\; \mathrm{V}$. The minimum fuse capacity (rated value) of the building will be: .......... $A$
A galvanometer of resistance, $G,$ is connected in a circuit. Now a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ is
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $B$ with respect to the point $A$ is$.......V$
The circuit diagram shown consists of a large number of element (each element has two resistors $R_1$ and $R_2$). The resistance of the resistors in each subsequent element differs by $a$ factor of $K = 1/2$ from the resistance of the resistors in the previous elements. The equivalent reistance between $A$ and $B$ shown in figure is :
Find the number of photons emitted per second from of source of light which results in a photocurrent with drift velocity of $1.5\ m/s$ in a conductor with cross-section area $0.25\ m^2$ , volume density of electrons $10^{20}\ per \ m^3$ , (Assume that $60\%$ of photons emitted result in electron emission)
A cell, shunted by a $8 \; \Omega$ resistance, is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega .$
Two conductors made of the same material are connected across a common potential difference. Conductor $A$ has twice the diameter and twice the length of conductor $B$. The power delivered to the two conductors ${P_A}$ and ${P_B}$ respectively is such that ${P_A}/{P_B}$ equals to