A filament bulb $(500 \,W,\,\, 100 \,V)$ is to be used in a $230\, V$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500\,W.$ The value of $R$ is .................. $\Omega$
NEET 2016, Medium
Download our app for free and get started
Resistance of bulb, $R_{B}=\frac{V^{2}}{P}=\frac{(100)^{2}}{500}=20\, \Omega$
Power of the bulb in the circuit,
$P =V I $
$I =\frac{P}{V_{B}} $
$=\frac{500}{100}=5\, \mathrm{A}$
$V_{R} =I R \Rightarrow(230-100)=5 \times R $
$\therefore \quad R =26 \,\Omega$
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
- 1For what value of $R$ the net resistance of the circuit will be $18\, ohms$ ............... $\Omega$View Solution
- 2In the circuit diagram shown in figure given below, the current flowing through resistance $3\, \Omega$ is $\frac{ x }{3}\,A$. The value of $x$ is $...........$View Solution
- 3View SolutionDiagram shows a circuit diagram choose incorrect statement
- 4In the ladder network shown, current through the resistor $3\, \Omega$ is $0.25\,A$. The input voltage $'V'$ is equal to:- ........... $V$View Solution
- 5A wire $50\, cm$ long and $1\;mm^2$ in cross -section carries a current of $4\; A$ when connected to a $2\; V$ battery. The resistivity of the wire isView Solution
- 6View SolutionOhm's law fails in
- 7Three resistors of $4\,\Omega ,6\,\Omega $ and $12\,\Omega $ are connected in parallel and the combination is connected in series with a $1.5\, V$ battery of $1\,\Omega $ internal resistance. The rate of Joule heating in the $4\,\Omega $ resistor is ................ $W$View Solution
- 8View SolutionThe relaxation time in conductors
- 9A uniform wire of length $l$ and radius $r$ has a resistance of $100\, \Omega $. It is recast into a wire of radius $\frac{r}{2}$. The resistance of new wire will be ............... $\Omega$View Solution
- 10A galvanometer of resistance, $G,$ is connected in a circuit. Now a resistance $R$ is connected in series of galvanometer. To keep the main current in the circuit unchanged, the resistance to be put in parallel with the series combination of $G$ and $R$ isView Solution