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
Discussion:
19 comments Page 2 of 2.
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.
R = R1+R2 R = V/I = 120/3 = 40.
R1+R2 = 20.
R2 = 40-20 = 20.
Ravi sankh said:
1 decade ago
If current & voltage doesn't change, then resistance also not change that's why answer is 20 ohm.
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.
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.
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.
Sridhar said:
1 decade ago
Can anyone say me why lamp resistance is specified here?
Anand said:
1 decade ago
Nice answer sugasini
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
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
Sibananda chakra said:
1 decade ago
Thank you. Friend.
Sugasini said:
1 decade ago
R=V/I
R1+R2=120/3=40
SO, R1=40-20=20
ANS 20
R1+R2=120/3=40
SO, R1=40-20=20
ANS 20
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