Two bulbs $A$ and $B$ are rated as $90\, W -120 \,V$ and $60\, W -120\, V$ respectively. They are connected in parallel across a $120\, V$ source. Find the current in each bulb. Which bulb will consume more energy ?
Two bulbs $A$ and $B$ are rated as $90\, W -120 \,V$ and $60\, W -120\, V$ respectively. They are connected in parallel across a $120\, V$ source. Find the current in each bulb. Which bulb will consume more energy ?
$(i)$ For the first bulb,
Resistance, $R_{1}=\frac{V^{2}}{P_{1}}=\frac{(120)^{2}}{90}=160 \Omega$
$\therefore \quad$ Current, $I_{1}=\frac{V}{R}=\frac{120}{160}=0.75 A$
For the second bulb,
Resistance, $R _{2}=\frac{ V ^{2}}{ P _{2}}=\frac{(120)^{2}}{60}=240 \Omega$
$\therefore \quad$ Current $, I _{2}=\frac{ V }{ R }=\frac{120}{240}=0.5 A$
$(ii)$ Bulb $A$ will consume more energy, because it has more power.
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$(i)$ The quantity of charge flowing past a point multiplied by time is current.
$(ii)$ The flow of charge through a conducting wire connected to a cell is the result of chemical reaction inside the cell
$(iii)$ The resistivity of all pure metals increases with increase in temperature.
$(iv)$ Ohm's law is a relation between the power used in a circuit to the current and the potential difference.
$(v)$ $A$ series circuit has only one conducting path for the electrons that move through it; a parallel circuit has multiple conducting paths.
$(vi)$ A conducting wire offers resistance to the flow of electrons because electrons repel each other in the wire.
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The SI unit of electric charge is $.........$