Find the current in the $6\,\Omega$ resistance.

$A$ cell whose e.m.f. is $2\, V$ and internal resistance is $0.1\,\Omega$ is connected with a resistance of $3.9\,\Omega$. The voltage across the cell terminal will be ................ $V$.

In the circuit shown in the figure,the resistance of the voltmeter is $6 \, k\Omega$. The voltmeter reading will be ................. $V$.

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 $\delta R_P \approx \delta R_Q$?

$A, B$ and $C$ are voltmeters of resistance $R, 1.5R$ and $3R$ respectively as shown in the figure. When some potential difference is applied between $X$ and $Y$,the voltmeter readings are $V_A, V_B$ and $V_C$ respectively. Then

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