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 1 of 2.
Raji said:
1 decade ago
The fuse rating is 3A. So without blowing the fuse the maximum current can be flown through the circuit is 3A.
When I = 3A, source voltage = 120v.
Resistance = 120/3 = 40 ohms.
So the resistance can be applied from rheostat without damaging fuse = 40-20 = 20 ohms.
When I = 3A, source voltage = 120v.
Resistance = 120/3 = 40 ohms.
So the resistance can be applied from rheostat without damaging fuse = 40-20 = 20 ohms.
(2)
Shivam sahu said:
9 years ago
As resistance is inversely proportional to current. So to calculate minimum resistance we have to consider max. Current through the circuit. So, I=3 A and with the help of this we can find answer as per the mentioned solutions.
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.
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.
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.
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.
Solayman Hayder said:
9 years ago
We know that, V=IR (ohms law)
Here,
v = 120v.
I = 3A, fuse will work at 3A without burning,
R1= 20ohm, R2=?
R1 + R2 = (120/3) = 40,
R2 = 40 - 20 = 20 ohms (Ans.)
Here,
v = 120v.
I = 3A, fuse will work at 3A without burning,
R1= 20ohm, R2=?
R1 + R2 = (120/3) = 40,
R2 = 40 - 20 = 20 ohms (Ans.)
(6)
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
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.
Sridhar said:
1 decade ago
Can anyone say me why lamp resistance is specified here?
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