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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