Current Electricity Test

Result

Current Electricity Test - 2

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Question 1
4 / -1

Ten million electrons pass from point P to point Q in one micro second. The current and its direction is

A
1.6×10⁻¹⁴A, from point P to point Q

B
3.2×10⁻¹⁴A, from point P to point Q

C
1.6×10⁻⁶A, from point Q to point P

D
3.2×10⁻¹²A, from point Q to point P

Solution

Here,number of electron,
n = 10000000 = 10⁷
Total charge on ten million electrons is
Q = ne [where e = 1.6 × 10⁻¹⁹C]
=10⁷× 1.6 × 10⁻¹⁹C = 1.6 × 10⁻¹²C
Time taken by ten million electrons to pass from point P to point Q is
t = 1μs = 10⁻⁶s
The current
I =
Since the direction of the current is always opposite to the direction of flow of electrons. Therefore due to flow of electrons from point P to point Q the current will flow from point Q to point P.
Question 2
4 / -1

1 ampere current is equivalent to

A
6.25×10¹⁸ electrons s⁻¹

B
2.25×10¹⁸ electrons s⁻¹

C
6.25×10¹⁴ electrons s⁻¹

D
2.25×10¹⁴ electrons s⁻¹

Solution

Q = It also Q = ne [e = 1.6 × 10⁻¹⁹C]
∴ ne = It or
= 6.25 × 10¹⁸ electrons s⁻¹
Question 3
4 / -1

A current in a wire is given by the equation, I =2t² − 3t + 1, the charge through cross section of wire in time interval t = 3s to t = 5s is

A
32.33C

B
43.34C

C
45.5C

D
42C

Solution

As I = dQ/dt ;
dQ − Idt ;
dQ = (2t²− 3t + 1)dt

Q.=
= [2/5(125 − 27) − 32(25 − 9) + 2]
= 43.34C
Question 4
4 / -1

The current in a wire varies with time according to the equation I = 4 + 2t, where I is in ampere and t is in second. The quantity of charge which has to be passed through a cross-section of the wire during the time t = 2s to t = 6s is?

A
40 C

B
48 C

C
38 C

D
43 C

Solution

Given:
I = 4 + 2t
Let dq be the charge which passes in a small interval of time dt. Then
dq = Idt
or dq = (4 + 2t) dt
On integrating, we get

= 48 C
Question 5
4 / -1

In an atom electrons revolves around the nucleus along a path of radius 0.72Å making 9.4×10¹⁸ revolution per second. The equivalent current is (e = 1.6×10⁻¹⁹C)

A
1.2A

B
1.5A

C
1.4A

D
1.8A

Solution

Radius of electron orbit
r = 0.72Å = 0.72 × 10⁻¹⁰m
Frequency of revolution of electron in orbit of given atom
v = 9.4 × 10¹⁸ rev/s
(where T is time period of revolution of electron in orbit)
∴ Then equivalent current is
I = e/T = ev = 1.6 × 10⁻¹⁹× 9.4 × 10¹⁸
= 1.504A