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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