When electric bulbs of same power,but different marked voltage are connected in series across the power line,their brightness will be:

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 $\delta R_P \approx \delta R_Q$?

In the given circuit,the value of $\left|\frac{I_1+I_3}{I_2}\right|$ is

$A$ uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop $(ABCD)$ (see figure). $A$ resistance of $2 \ \Omega$ is connected between points $B$ and $D$ and a battery of $2 \ V$ is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(I)$ is: (in $A$)

In the adjacent circuit,a voltmeter of internal resistance $R$,when connected across $B$ and $C$,reads $\frac{100}{3} \text{ V}$. Neglecting the internal resistance of the battery,the value of $R$ is:

The internal resistances of two cells shown in the figure are $0.1\,\Omega$ and $0.3\,\Omega$. If $R = 0.2\,\Omega$,find the potential difference across the cells.