In the circuit shown in the figure,the power developed across the $1 \Omega$,$2 \Omega$,and $3 \Omega$ resistances are in the ratio:

$A$ constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

$A$ 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 $A$)

Find the potential of $J$ with respect to $G$ in $V$.

Resistances $R_1$ and $R_2$ are joined in parallel and a current is passed so that the amount of heat liberated is $H_1$ and $H_2$ respectively. The ratio $\frac{H_1}{H_2}$ has the value:

As shown in the figure,the voltmeter reads $2\,V$ across the $5\,\Omega$ resistor. The resistance of the voltmeter is $.......\,\Omega$.