The ratio of powers dissipated respectively in $R$ and $3R$,as shown in the figure,is:

$A$ uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop $(ABCD)$ (see figure). $A$ resistance of $2 \ \Omega$ is connected between points $B$ and $D$ and a battery of $2 \ V$ is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(I)$ is: (in $A$)

$A$ resistor ${R_1}$ dissipates the power $P$ when connected to a certain generator. If the resistor ${R_2}$ is put in series with ${R_1}$,the power dissipated by ${R_1}$

Electric bulbs rated $50\, W$ - $100\, V$ glowing at full power are to be used in parallel with a battery of $120\, V$ and internal resistance $10\,\Omega$. The maximum number of bulbs that can be connected so that they glow at full power is:

Two cells of same $emf$ $E$ but internal resistances $r_1$ and $r_2$ are connected in series to an external resistor $R$ (as shown in the figure). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero?

$A$ 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.