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Lakhmir Singh & Manjit Kaur - Physics

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Q. 34.2( 31 Votes )

An electric bulb

Class 10th Lakhmir Singh & Manjit Kaur - Physics 1. Electricity

Answer :

(a) Here resistance are connected in series

Total resistance of the circuit = R₁+R₂=20+4=24Ω(ohm)

(b) According to ohm’s law.

V=IR

Therefore,

6V=I x 24Ω(ohm)

I=6 V /24 ohm = 0.25A(Ampere)

(c) Potential difference across the electric bulb

V₁= IR₁=0.25 X 20=5V

(d) Potential difference across the resistance wire

V₂= IR₂=0.25X4=1V

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