The terminal voltage of the battery, whose emf is $10 \, V$ and internal resistance is $1 \, \Omega$, when connected through an external resistance of $4 \, \Omega$ as shown in the figure is: (in $V$)

$A$ galvanometer of resistance $G$ is connected in a circuit. Now,a resistance $R$ is connected in series with the galvanometer. To keep the main current in the circuit unchanged,the resistance $S$ to be put in parallel with the series combination of $G$ and $R$ is:

In the given circuit,the terminal potential difference of the cell is: (in $ V$)

In the given electric circuit, the potential difference across $PQ$ is approximately ...... volts.

In the given circuit,the internal resistance of the cell is zero. If $i_1$ and $i_2$ are the readings of the ammeter when the key $(K)$ is opened and closed respectively,then $i_1: i_2=$

For the given circuit,in the steady state,$\left| V_{B}-V_{D}\right| = .......V.$