Electrical Engineering - Ohm's Law - Discussion

Discussion Forum : Ohm's Law - General Questions (Q.No. 23)
23.
A 120 V lamp-dimming circuit is controlled by a rheostat and protected from excessive current by a 3 A fuse. To what minimum resistance value can the rheostat be set without blowing the fuse? Assume a lamp resistance of 20 ohms.
40
4
2
20
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 1 of 2.

Sugasini said:   1 decade ago
R=V/I
R1+R2=120/3=40
SO, R1=40-20=20
ANS 20

Sibananda chakra said:   1 decade ago
Thank you. Friend.

Harpreet said:   1 decade ago
By ohm law
V=IR ---->1
SO
V=120v
and I=3A
R1=20
R2=?
SO ALL THE VALUES PUT IN ONE
WE GET
V=I(R1+R2)
120 = 3 (20+R2)
R2=20 ohms

Anand said:   1 decade ago
Nice answer sugasini

Sridhar said:   1 decade ago
Can anyone say me why lamp resistance is specified here?

Krishna said:   1 decade ago
If lamp resistance of 20 ohm is not given then obviously the min resistance would have been as 40 ohms for this question, since the lamp also will be having some resistance by considering that one we got the ans as 20 ohms.

Darshanmaths said:   1 decade ago
V1=120V.

I1=3A.

Lamp resistance=20 ohms.

Total resistance=V1/R1=120/3=40 ohms.

As we know,

Total resistance=Lamp resistance + Resistance of rheostat.

Hence
Resistance of rheostat=20 ohms.

Ravi sankh said:   1 decade ago
If current & voltage doesn't change, then resistance also not change that's why answer is 20 ohm.

Mallikarjunreddy said:   1 decade ago
V = IR we know (r1= lamp resistance = 20 ohms, I = 3A).

R = R1+R2 R = V/I = 120/3 = 40.

R1+R2 = 20.

R2 = 40-20 = 20.

Souvik said:   1 decade ago
We have V=120v and Critical current as 3A.

So the circuit resistance has to be- 120/3 = 40 Ohms.

We have a bulb of 20 Ohms. Hence the rheostat can be set to a minimum of= 40-20 = 20 Ohms.

Answer = 20 Ohms.


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