Four identical bulbs,each rated $100 \, W, 220 \, V$,are connected across a battery as shown. The total electric power consumed by the bulbs is.......$W$.

The equivalent resistance of a group of resistances is $R$. If another resistance is connected in parallel to the group,its new equivalent becomes $R_1$,and if it is connected in series to the group,its new equivalent becomes $R_2$. We have:

The ammeter reading in the circuit below is .............. $A$.

$A$ $100\, W$ bulb $B_1$ and two $60\, W$ bulbs $B_2$ and $B_3$ are connected to a $220\, V$ source,as shown in the figure. If $P_1, P_2$,and $P_3$ are the output powers of the bulbs $B_1, B_2$,and $B_3$ respectively,then:

$A$ wire of resistance $20 \Omega$ is divided into $10$ equal parts. $A$ combination of two parts are connected in parallel and so on. Now,the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is . . . . . . $\Omega$.

Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square (in $\Omega$)?