Two resistances are joined in parallel whose resultant is $6/8 \, \Omega$. One of the resistance wires is broken and the effective resistance becomes $2 \, \Omega$. Then the resistance in ohm of the wire that got broken was:

$A$ ring is made of a wire having a total resistance $R_0 = 12\,\Omega$. Find the ratio of the lengths $\frac{\ell_1}{\ell_2}$ of the two arcs between points $A$ and $B$ as shown in the figure,such that the equivalent resistance $R$ of the circuit between these points is $\frac{8}{3}\,\Omega$.

In the given circuit,the current flowing through the resistance $20 \ \Omega$ is $0.3 \ A$,while the ammeter reads $0.9 \ A$. The value of $R_1$ is . . . . . . $\Omega$.

Three voltmeters $A$,$B$ and $C$ having resistances $R$,$1.5 R$ and $3 R$ respectively are used in a circuit as shown. When a potential difference is applied between $X$ and $Y$,the readings of the voltmeters are $V_{1}$,$V_{2}$,and $V_{3}$ respectively. Then

$A$ wire of resistance $12 \,\Omega$ is bent into a circle. What is the equivalent resistance between two diametrically opposite points in $\Omega$?

If in the circuit,power dissipation is $150\, W$,then $R$ is ............... $\Omega$.