Each of the resistances in the network shown in the figure is equal to $R$. The resistance between the terminals $A$ and $B$ is

Two resistances of $400 \Omega$ and $800 \Omega$ are connected in series with a $6 \text{ V}$ battery of negligible internal resistance. $A$ voltmeter of resistance $10000 \Omega$ is used to measure the potential difference across the $400 \Omega$ resistor. The error in the measurement of potential difference in volts is approximately:

Six equal resistances $R$ are connected between points $P$,$Q$,and $R$ as shown in the figure. Then the net resistance will be maximum between:

Calculate the current shown by the ammeter $A$ in the circuit diagram. (in $A$)

In the given circuit,the potential of the point $E$ is

Two equal resistances, $400 \Omega$ each, are connected in series with an $8 V$ battery. If the resistance of the first one increases by $0.5 \%$, what is the change required in the resistance of the second one in order to keep the potential difference across it unaltered?