Calculate the amount of charge flowing in $2$ minutes in a wire of resistance $10\,\Omega$ when a potential difference of $20\,V$ is applied between its ends.

$A$ $dc$ source of $emf \, E_1 = 100 \, V$ and internal resistance $r = 0.5 \, \Omega$,a storage battery of $emf \, E_2 = 90 \, V$,and an external resistance $R$ are connected as shown in the figure. For what value of $R$ will no current pass through the battery $E_2$ (in $.5$)? ................ $\Omega$

$A$ battery of internal resistance $4 \, \Omega$ is connected to the network of resistors as shown. In order for the maximum power to be delivered to the network, the value of $R$ in $\Omega$ should be:

$A$ set of $n$ equal resistors,each of value $R$,are connected in series to a battery of emf $E$ and internal resistance $R$. The current drawn is $I$. Now,the $n$ resistors are connected in parallel to the same battery. The current drawn from the battery becomes $10I$. The value of $n$ is:

What will be the equivalent resistance between the two points $A$ and $D$ in the given circuit (in $\Omega$)? (All resistors are $10 \ \Omega$ each.)

$R_1$ and $R_2$ are two resistors made of different materials. The temperature coefficient of resistance for $R_1$ is $\alpha$ and for $R_2$ is $-\beta$. If the resistance of the series combination of $R_1$ and $R_2$ does not change with temperature,then the ratio of the resistances of the two wires is: