Physics · Current Electricity
Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive practice and topic analytics.
When a potential difference is applied across the ends of a uniform wire of length l and radius r, a current I flows in the wire. If same potential difference is applied to the ends of another wire of the same material but of length 2l and radius 2r, the current in the wire is:
- A
I/4
- B
2I
- C
I
- D
I/2
R=ρl/A
When radius is doubled : Area=2*pi*(2r) = 2*pi*4r
Hence the area has increased by 4 times.
New resistance : Rx = ρ*2l/4A = ½ R
V=IR where V remains same and R becomes half
I = V / ½R
I become equal to 2I.
R=pl/A
When radius is doubled:
Area=2π(2r) = 2π4r
Hence the area has increased by 4 times.
New resistance: R'= p2l/4A =1/2 R
V=IR
where V remains same and R becomes half
I = V / ½R
I became 2I.
R=pl/A
When radius is doubled:
Area=2π(2r) = 2π4r
Hence the area has increased by 4 times.
New resistance: R'= p2l/4A =1/2 R
V=IR
where V remains same and R becomes half
I = V / ½R
I became 2I.
R=pl/A
When radius is doubled:
Area=2π(2r) = 2π4r
Hence the area has increased by 4 times.
New resistance: R'= p2l/4A =1/2 R
V=IR
where V remains same and R becomes half
I = V / ½R
I became 2I.
R=pl/A
When radius is doubled:
Area=2π(2r) = 2π4r
Hence the area has increased by 4 times.
New resistance: R'= p2l/4A =1/2 R
V=IR
where V remains same and R becomes half
I = V / ½R
I became 2I.
Tagged under Physics · Current Electricity · 2018