In the circuit diagram shown below,the magnitude and direction of the flow of current respectively would be

In the circuit given here,the potentials at points $A$,$B$,and $C$ are $70 \,V$,$0 \,V$,and $10 \,V$ respectively. Then:

The resistance of each arm of the Wheatstone bridge is $10 \, \Omega$. A resistance of $10 \, \Omega$ is connected in series with a galvanometer. Then, the equivalent resistance across the battery will be:

In the circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $Omega$)

$A$ Wheatstone bridge is initially at room temperature and all arms of the bridge have the same value of resistance $(R_1=R_2=R_3=R_4=R)$. When $R_3$ is heated to some temperature,its resistance value increases by $10 \%$. The potential difference $(V_a - V_b)$ (after $R_3$ is heated) is . . . . . . $V$. Given the total voltage is $40 \text{ V}$.

The four arms of a Wheatstone bridge have resistances as shown in the figure. $A$ galvanometer of $15\, \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10\, V$ is maintained across $AC$.