In the given circuit diagram,the galvanometer reading is zero. If the internal resistance of the cell is neglected,then the value of $X$ is ......... $\Omega$.

$A$ cell of emf $90\,V$ is connected across a series combination of two resistors,each of $100\,\Omega$ resistance. $A$ voltmeter of resistance $400\,\Omega$ is used to measure the potential difference across one of the resistors. The reading of the voltmeter will be $.........\,V$.

The magnitude of $i$ in ampere unit is

The equivalent resistance between $x$ and $y$ in the circuit shown is ............. $\Omega$

Consider the two circuits $P$ and $Q$ shown below,which are used to measure the unknown resistance $R$. In each case,the resistance is estimated by using Ohm's law $R_{\text{est}} = \frac{V}{I}$,where $V$ and $I$ are the readings of the voltmeter and the ammeter,respectively. The meter resistances $R_V$ and $R_A$ are such that $R_A \ll R \ll R_V$. The internal resistance of the battery may be ignored. The absolute error in the estimate of the resistance is denoted by $\delta R = |R - R_{\text{est}}|$.
$(a)$ Express $\delta R_P$ in terms of the given resistance values.
$(b)$ Express $\delta R_Q$ in terms of the given resistance values.
$(c)$ For what value of $R$ will $\delta R_P \approx \delta R_Q$?

$A$ battery with an internal resistance of $4\,\Omega$ is connected to a circuit as shown in the figure. For the circuit to receive maximum power,the value of $R$ should be ..... $\Omega$.