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.

Singh said:   3 years ago
Thank you.

Vikas said:   7 years ago
Thanks a lot for explaining the solution.

Suresh said:   8 years ago
Thank you all.

Victor said:   8 years ago
Thank you @Darshan.
(1)

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.

Jagadish said:   9 years ago
I got it, Thanks @Raji.

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.)
(6)

Sadbi Ahmad said:   9 years ago
Thank you everybody for solving the solution.

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.
(2)

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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