In the given circuit,current $I$ is independent of the resistance $R_6$. Then

Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is: (in $\Omega$)

In the given figure: $V_1=V, V_2=\alpha V, R_1=\beta R, R_2=\gamma R$,where $\alpha, \beta$,and $\gamma$ are positive real numbers. The value of current $I$ is

The current drawn from the battery in the given network is (Internal resistance of the battery is negligible). (in $A$)

Four ammeters with identical internal resistances $r$ and a resistor of resistance $R$ are connected to a current source as shown in the figure. It is known that the reading of the ammeter $A_1$ is $I_1 = 3\ A$ and the reading of the ammeter $A_2$ is $I_2 = 5\ A$. Determine the ratio of the resistances $R/r$.

The current through the $1 \Omega$ resistance in the following circuit is (in $A$)