$A$ car has a fresh storage battery of emf $12 \, V$ and internal resistance $2 \times 10^{-2} \, \Omega$. If the starter motor draws a current of $80 \, A$, then the terminal voltage when the starter is $ON$ is: (in $V$)

$A$ $16 \ \Omega$ wire is bent to form a square loop. $A$ $9 \ V$ battery with internal resistance $1 \ \Omega$ is connected across one of its sides. If a $4 \ \mu F$ capacitor is connected across one of its diagonals,the energy stored by the capacitor will be $\frac{x}{2} \ \mu J$,where $x = . . . . . . .$

$A$ light bulb of resistance $R = 16 \, \Omega$ is attached in series with an infinite resistor network with identical resistances $r$ as shown below. $A$ $10 \, V$ battery drives current in the circuit. Find the value of $r$ in $\Omega$ such that the bulb dissipates about $1 \, W$ of power.

The potential difference between the points $P$ and $Q$ is nearly: (in $\,V$)

As shown,the circuit is made of $8$ different resistors. It is found that when $R_1 = 4\,\,\Omega,$ the resistance between $A$ and $B$ is $2\,\,\Omega.$ Now replace $R_1$ by a $6\,\,\Omega$ resistor,what is the resistance between $A$ and $B$?

All wires have the same resistance $r$. The equivalent resistance between $A$ and $B$ is $R$. Now,if the keys $K_1$ and $K_2$ are closed,what will be the new equivalent resistance?