$A$ bulb rated $30\,V, 90\,W$ is to be operated on a $120\,V$ supply. What resistance (in $\Omega$) must be connected in series with it?

In the following circuit diagram,$E = 4\,V, r = 1\,\Omega$ and $R = 45\, \Omega$,then the reading of the ammeter $A$ will be

Two resistors of $200\Omega$ and $400\Omega$ are connected in series with a battery of $100\text{ V}$. $A$ bulb rated at $200\text{ V}, 100\text{ W}$ is connected across the $400\Omega$ resistance. The potential drop across the bulb is . . . . . . $\text{V}$.

In the given circuit,if the current flowing through the $5 \Omega$ resistor is $0.5 \text{ A}$,then the value of $E$ is: (in $\text{ V}$)

The equivalent resistance between points $A$ and $B$ of an infinite network of resistances each of $1\,\Omega$ connected as shown,is

If $E_1 = 4 \ V$ and $E_2 = 12 \ V$,find the current in the circuit and the potential difference between the points $P$ and $Q$ respectively.