In the circuit shown,the cells $A$ and $B$ have negligible resistance. For $V_{A} = 12 \; V$,$R_{1} = 500 \; \Omega$ and $R = 100 \; \Omega$,the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$.

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 $\delta R_P \approx \delta R_Q$?

The charge flowing through a resistor $R$ varies with time $t$ as $Q = 3t - 6t^2$. The heat produced in $R$ until the current in it becomes zero is:

The potential difference between points $A$ and $B$ of the adjoining figure is

$A$ battery of $e.m.f.$ $10\, V$ and internal resistance $0.5\, \Omega$ is connected across a variable resistance $R$. The value of $R$ for which the power delivered in it is maximum is given by ......... $\Omega$.

In the circuit shown,the capacitance of the capacitor is $2\,\mu F$. Find the current flowing through the $2\,\Omega$ resistor in $A$.