An electric bulb of resistance 20Ω and a resistance wire of 4 Ω are connected in series with a 6 V battery. Draw the circuit diagram and calculate:

(a) total resistance of the circuit


(b) current through the circuit.


(c) potential difference across the electric bulb.


(d) potential difference across the resistance wire.



(a) Here resistance are connected in series


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


(b) According to ohm’s law.


V=IR


Therefore,


6V=I x 24 ohm


I=6 V /24 ohm = 0.25amp


(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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