A Copper $(Cu)$ rod of length $25\, {cm}$ and cross- sectional area $3\, {mm}^{2}$ is joined with a similar Aluminium $(Al)$ rod as shown in figure. Find the resistance of the combination between the ends $A$ and $B$ (in ${m} \Omega$)
(Take Resistivity of Copper $=1.7 \times 10^{-8}\, \Omega \,{m}$, Resistivity of Aluminium $=2.6 \times 10^{-8}\, \Omega \,{m}$ )
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.*
Resistances ${R_1}$ and ${R_2}$ are joined in parallel and a current is passed so that the amount of heat liberated is ${H_1}$ and ${H_2}$ respectively. The ratio $\frac{{{H_1}}}{{{H_2}}}$ has the value
Under what condition current passing through the resistance $R$ can be increased by short circuiting the battery of emf $E_2$. The internal resistances of the two batteries are $r_1$ and $r_2$ respectively.
Two cells are connected in opposition as shown. Cell $\mathrm{E}_1$ is of $8 \mathrm{~V}$ emf and $2 \ \Omega$ internal resistance; the cell $E_2$ is of $2 \mathrm{~V}$ emf and $4\ \Omega$ internal resistance. The terminal potential difference of cell $\mathrm{E}_2$ is:
The resistive network shown below is connected to a $D.C.$ source of $16\, V$. The power consumed by the network is $4\, Watt$. The value of $R$ is ............. $\Omega$
In potentiometer experiment when $K_1$ is closed balance length is $100\,cm$. Then what will be balancing length when $K_2$ is closed ................ $\mathrm{cm}$
$12$ 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 :-
A Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\, ohm$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $ohm$
For a cell terminal $P.D.$ is $2.2\;V$ when circuit is open and reduces to $1.8\;V$ when cell is connected to a resistance of $R = 5\,\Omega $. Determine internal resistance of cell $(r)$ is then ........ $\Omega$