NEET-XII-Physics
32: Electric Current in Conductors
- #27A bulb is made using two filaments. A switch selects whether the filaments are used individually or in parallel. When used with a 15 V battery, the bulb can be operated at 5 W, 10 W or 15 W. What should be the resistances of the filaments?Ans : We know:
`` R=\frac{{V}^{2}}{P}``,
where R is the resistance, V is the voltage drop and P is the power on which the bulb is operated.
We can calculate the resistance of the bulb for the same V and different P.
The resistances of the bulb for three different powers are:
`` {R}_{1}=\frac{{\left(15\right)}^{2}}{5}=45\,\mathrm{\,\Omega \,}``
`` {R}_{2}=\frac{{\left(15\right)}^{2}}{10}=22.5\,\mathrm{\,\Omega \,}``
`` {R}_{3}=\frac{{\left(15\right)}^{2}}{15}=15\,\mathrm{\,\Omega \,}``
When the two resistance are used in parallel, the equivalent resistance will be less than their individual resistance.
∴ The two resistance are 45 Ω and 22.5 Ω.
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