Question

An electric bulb of resistance 20 ohm and a resistance wire of 4 ohm are connected in series with a 6 v battery .Draw the circuit diagram and calculate 1)total resistance of the circuitb
2)current through the circuitc.
3)P.D across the electric bulbd.
4) p.d across the resistance wire

Answer 1

see circuit diagram,
both wire and bulb are in series.
so, equivalent resistance=6+20= 24 ohm'
so, 1) 24 ohm'

2) current through circuit= v/r = 6/24 = 0.25 ohm'

3)PD across bulb = i×r(bulb) = 0.25× 20= 5 volt

4)PD across wire= i× r( wire)= 0.25× 4= 1 volt.



YOUR ANSWER

Answer 2

Answer:

Given

Battery=6v

R1=20 ohm

R2=4ohm

We know that,

(a) total resistence=R1+R2

=20+4

=24ohm

(b) current in circuit= v/r

=6/24=0.25A

(c) PD across bulb=0.25×20

=5v

(D) PD across wire= 0.25×4

=1v

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