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

Two resistors of $100 \Omega$ each are connected in series with a $9 \text{ V}$ battery. $A$ voltmeter of $400 \Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is . . . . . . $V$.

In the circuit shown in the figure, if the ammeter and voltmeter are ideal, then the power consumed in the $9 \, \Omega$ resistor will be .......... $W$.

In terms of resistance $R$ and time $T$,the dimensions of the ratio $\frac{\mu}{\varepsilon}$ of the permeability $\mu$ and permittivity $\varepsilon$ is:

The series combination of two batteries,both of the same emf $10 \; V,$ but different internal resistances of $20 \; \Omega$ and $5 \; \Omega,$ is connected to the parallel combination of two resistors $30 \; \Omega$ and $R \; \Omega.$ If the voltage difference across the battery of internal resistance $20 \; \Omega$ is zero,the value of $R$ (in $\Omega$) is:

$A$ simple circuit contains an ideal battery and a resistance $R$. If a second resistor is placed in parallel with the first,which of the following statements is correct?