When electric bulbs of same power, but different marked voltage are connected in series across the power line, their brightness will be :
Medium
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1In the diagram resistance between any two junctions is $R$. Equivalent resistance across terminals $A$ and $B$ isView Solution
- 2For comparing the $e.m.f.$'s of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessfulView Solution
- 3Two sources of equal $emf$ are connected to an external resistance $R$. The internal resistances of the two sources are ${R_1}$ and ${R_2}\,({R_2} > {R_1})$. If the potential difference across the source having internal resistance ${R_2}$ is zero, thenView Solution
- 4The equivalent resistance between $A$ and $B$ of the network shown in figureView Solution
- 5Resistance in the two gaps of a meter bridge are $10\, ohm$ and $30\, ohm$ respectively. If the resistances are interchanged the balance point shifts by.............. $cm$View Solution
- 6In the figure, the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of $\mathrm{emf}$ $\varepsilon$ and internal resistance $r$. The cell $C’s$ $\mathrm{emf}$ is $\varepsilon /2$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ isView Solution
- 7A wire when connected to $220\,V$ mains supply has power dissipation ${P_1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is ${P_2}$. Then ${P_2}:{P_1}$ isView Solution
- 8During lighting, a current pulse, shown in figure, flows from the cloud at a height $1.5\ km$ to the ground. If the breakdown electric field of humid air is about $400\ kVm^{-1}$ , the energy released during lighting would be (in unit of $10^9\ J$ )View Solution
- 9In the circuit here, the steady state voltage across capacitor $C$ is a fraction of the battery $e.m.f.$ The fraction is decided byView Solution
- 10On an electric heater $220\, volt$ and $1100\, watt$ are marked. On using it for $4$ hours, the energy consumed in $kWh$ will beView Solution