Question

an electric lamp ad a conductor of resistance 4 ohm are connected in series to a 6 volt battery . current drawn by lamp is 0.35 Ampere find the resistance of electric lamp​

Answer 1

Solution :
Lamp :

Resistance , R1=?
R
1
=
?

Potential differences, V1=?
V
1
=
?

Current drawn, I1=0.25A
I
1
=
0.25
A

Other conductor :
:

Resistance , R2=4Ω
R
2
=
4
Ω

Potential differences , V2=?
V
2
=
?

Current drawn , I2=?
I
2
=
?

Resultant resistance, R=R1+R2
R
=
R
1
+
R
2

∴
∴
Potential differene, V=6V
V
=
6
V

Current in the circuit , I=0.25A
I
=
0.25
A

∵I=I1=I2
∵
I
=
I
1
=
I
2

∴V=IR
∴
V
=
I
R

⇒R=VI=60.25=6×100425=24Ω
⇒
R
=
V
I
=
6
0.25
=
6
×
100
4
25
=
24
Ω

Now, R1+R2=R
R
1
+
R
2
=
R
⇒R1+4=24
⇒
R
1
+
4
=
24

∴R1=24−4=20Ω
∴
R
1
=
24
-
4
=
20
Ω





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Answer 2

Answer:

20 ohm

Explanation:

since the current in series is equal

therefore, the current through 4 ohm resistance will also be 0.25 A.

now,

voltage of the conductor = 4×0.25

= 1 V

therefore voltage of bulb = total voltage - voltage of conductor

6 V - 1 V

= 5 V

now, resitance of the lamp = 5/0.25

= 20 ohm