MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Current Electricity Mcq
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
Carbon resistor used in electronic circuit are marked for their resistance value and tolerance by a colour scheme. A given resistor has a colour scheme brown, green and gold. It value is Ohms
0%
a)3.2×105 ± 5%
0%
b) 1.0×106 ± 10%
0%
c)1.0 ×106 ± 5%
0%
d)1.0×103 ± 5%
Explanation
Answer: (c)
Q.2
Through an electrolyte an electric current is due to drift of
0%
a) free electrons
0%
b) free electrons and holes
0%
c)positive and negative ions
0%
d)protons
Explanation
Answer:(c)
Q.3
The specific resistance of any known metal is not affected by change in
0%
a) temperature
0%
b) pressure
0%
c) applied magnetic field
0%
d) dimensions
Explanation
Answer: (d)
Q.4
A flow of 107 electrons per second in a conduction wire constitutes current of
0%
a)1.6×10-26 amp
0%
b) 1.6×1012 amp
0%
c)1.6×10-12 amp
0%
d)1.6×1026 amp
Explanation
Answer: (c)
Q.5
In the given figure equivalent resistance between P and Q will be
0%
a) 14/9Ω
0%
b) 9/14 Ω
0%
c) 14/3 Ω
0%
d) 3/14 Ω
Explanation
Given figure is Wheatstone bridge.Condition for balanced bridge is ,current will not flow through branch RS Hence 3 Ω, 4 Ω resistance are in series R’=3 Ω +4 Ω=7 Ω 6 Ω , 8 Ω are in series R”=6 Ω +8 Ω=14 ΩR’ and R” are connected in parallel let equivalent resistance be Reqon substitutions we getReq=14/3 Ω Answer: (c)
Q.6
If specific resistance of a potentiometer wire is 10-7 Ωm, the current flow through it is 0.1 A and the cross-sectional area of wire is 10-6 m2 then potential gradient will be [ CBSE PMT 2001]
0%
a)10-2 V/m
0%
b) 10-4 V/m
0%
c) 10-6 V/m
0%
d)10-8 V/m
Explanation
Potential gradient=Potential fall per unit length In this case resistance of unit length R=ρl / A=(10-7×1) / 10-6=0.1 m Potential fall across R is V=IR=0.1 ×0.1=0.01 V/m=10-2Answer: (a)
Q.7
An electron is revolving n times per second. The charge passing in t sec is
0%
a) net
0%
b) ne/t
0%
c)nt/e
0%
d)et/n
Explanation
Answer: (a)
Q.8
A resistance wire connected in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 :If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is : [ NEET 2020]
0%
a) 1.0×10−2 m
0%
b) 1.0×10−1 m
0%
c) 1.5×10−1 m
0%
d) 1.5×10−2 m
Explanation
Length of wire of resistance R = 1.5 m, S = 10 Ω R = 15 Ω = 1.5 m 1 Ω = 1.0×10−1 m Ans : (b)
Q.9
The color code of a resistance is given below The values of resistance and tolerance, respectively, are [NEET 2020]
0%
a) 470 kΩ, 5%
0%
b) 47 kΩ, 10%
0%
c) 4.7 kΩ, 5%
0%
d) 470 Ω, 5%
Explanation
Answer : (d)
Q.10
The reading of ideal voltmeter in the circuit shown is
0%
a) 0.6V
0%
b) 0V
0%
c) 0.5V
0%
d) 0.4V
Explanation
Voltmeter is ideal , have infinite resistance no current flows through it Both the branches are parallel to 2V Resistance in each branch is 50Ω Thus current flowing through each branch is 2/50 = 0.04 A Potential at point A = 2 – ( 20×0.04) = 1.2V Answer : (d)
Q.11
The meter bridge shown in the balance position with P/Q=l1/l2 .
0%
a) Yes P/Q=(l2-l1)/(l2+l1 ).
0%
b) No, no null point
0%
c) Yes P/Q=l2/l1
0%
d) Yes P/Q=l1/l2 .
Explanation
From given, we have, The metre bridge is in a balanced position with Р/Q = l1/l2. When we interchange the positions of galvanometer and cell, the metre bridge will remain in a balanced position with the current drawn from the battery being zero. So, the meter bridge will work. As, P ∝ l1 and Q ∝ l2 the condition will remain the same as Р/Q = l1/l2 after interchanging the positions of galvanometer and cell. Answer: ( d )
Q.12
) In the circuits shown below, the readings of voltmeters and the ammeters will be
0%
a) V1 = V2 and i1 < i2
0%
b) V1 = V2 and i1 = i2
0%
c) V1 > V2 and i1 > i2
0%
d) V1 = V2 and i1 = i2
Explanation
Energy=VIt Here I is current in circuit=2At is time duration for which current flows=6 min=6x60=360secEnergy=1000JOn substituting values in equation 1000=(V) 2 x 360 on simplification we getV=1.388 Ans : (b)
Q.13
A carbon resistor of (47±4.7)kΩ is to be marked with rings of different colours for its identification. The colour code sequence
0%
a) Yellow – Green – Violet – Gold
0%
b) Yellow – Violet –Orange – Silver
0%
c) Violet – Yellow – Orange – Silver
0%
d) Green – Orange– Violet– Gold
Explanation
Answer : (b)
Q.14
A set of 'n' equal resistors, of value 'R' each are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then, the current draw from battery become 10I. The value of 'n' is
0%
a) 20
0%
b) 11
0%
c) 10
0%
d) 9
Explanation
Total resistance = nR+R = R(n+1) E =I R(n+1) Now n resistance are connected parallel Thus now total resistance is n/R + R E = 10I (R/n + R) Answer : (c)
Q.15
A battery consists of a variable number ‘n’ of identical cells ( having internal resistance ‘r’ each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n
0%
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.16
The following unbalanced Wheatstone Bridge is constructed. Calculate the output voltage across points C and D [AFMC 1999]
0%
a) 60 V
0%
b) 35 V
0%
c) 100V
0%
d) 50V
Explanation
For the first series arm, ACB For the second series arm, ADB The voltage across points C-D is given as: 60 -25 = 35V Answer: (b)
Q.17
A 2Ω resistor is connected in series with R Ω resistor. This combination is connected across a cell. When the potential difference across 2 Ω resistor is balanced on potentiometer wire, null point is obtained at length of 300cm. When the same procedure is repeated for R Ω resistor, null point is obtained at length 350cm, value of R is .. [AFMC 2000]
0%
a) 5
0%
b) 3.33
0%
c) 4.6
0%
d) 2.33
Explanation
It can be solved on the basis of meter bridge. Null point is obtained. resistivity of meter bridge wire be ρ/cm here R1 =2Ω R2=300 ρΩ , R3=R Ω, R4=350 ρ Ω Answer : (d)
Q.18
Three bulbs of 40W, 60W and 100W are arranged in series with 220V, which bulb has minimum resistance? [AFMC 2001]
0%
a) 100W
0%
b) 40W
0%
c) 60W
0%
d) equal in all bulbs
Explanation
Bulbs are arranged in series so current through each bulb is sameP=I2R P ∝ R hence lower is the power lower is the resistance Answer : (b)
Q.19
The measurement of voltmeter in the following circuit is .[AFMC 2001]
0%
a) 6.0V
0%
b) 4.0V
0%
c) 3.4V
0%
d)2.4V
Explanation
There no potential drop up to the point where voltmeter is used. So it will measure full 6Volts Answer: (a)
Q.20
The potentiometer consists of wire of length 4m and resistance 10 立. It is connected to a cell of e.m.f 2V. The potential difference per unit length of the wire will be
0%
a) 10 V/m
0%
b) 5 V/m
0%
c) 2 V/m
0%
d) 0.5 V/m
Explanation
potential difference per unit length is potential gradientFormula for Potential gradient=V/length of potentiometer wire=2/4=0.5 V/m Ans : (d)
Q.21
An electron revolves 6×1015 times per second in a circular orbit. The current in the loop is
0%
a) 0.96 mA
0%
b) 0.96 µA
0%
c)28.8 A
0%
d)none of these
Explanation
I=ef here f=6×1015 Answer:(a)
Q.22
A 100W, 200V bulb is connected to a 160V supply. The actual power consumption would be … [AFMC 2002]
0%
a) 185W
0%
b) 100W
0%
b) 54W
0%
b) 64W
Explanation
Resistance of the bulb power p=100W , Voltage V=200VBy substituting values in above equation R=400 ΩNow new potential V=160VEquation for power POn substituting the values P=64W Answer : (d)
Q.23
The practical unit of resistance is 立. and1 立 is equal to : [AFMC 2003]
0%
a) 1018emu
0%
b) 109emu
0%
c) 1015emu
0%
d) none of these
Explanation
1立=109esuAnswer: (b)
Q.24
Kirchoff’s I and II laws are based on conservation of [AFMC 2003]
0%
a) Energy and Charge respectively
0%
b) Charge and energy respectively
0%
c) mass and charge respectively
0%
d) none of these above
Explanation
Ist law states that the algebraic sum of charges at the junction is zero , hence it is based on conservation of charge IIndlaw states that algebraic sum of the potential in closed path is zero, hence it is based on conservation of energy Ans : Answer: (b)
Q.25
Ohms Law is not obeyed by : [AFMC 2003]
0%
a) electrolytes
0%
b) discharge tube
0%
c) vacuum tube
0%
d) all of the above
Explanation
Vacuum tube is non-ohmic device, like semiconductor Answer: (c)
Q.26
A wire is cut in to 4 pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was R, the resistance of the bundle will be: [AFMC 2005]
0%
a) R/4
0%
b) R/8
0%
c) R/16
0%
d) R/32
Explanation
By cutting the wire in 4 pieces length is ¼ of original.Now bundle is formed hence bundle’s area becomes 4 times of original areaFormula for resistance is Now New length l’=l/4 a’=4aNew resistance R'OR each resistance is R/4 and connected in parallelHence R’=1/4 R/4=R/16Answer: (c)
Q.27
In the circuit shown in the figure the potential difference between X and Y will be [AFMC 2007]
0%
a) zero volts
0%
b) 20V
0%
c) 60V
0%
d) 120V
Explanation
At point X ,Y circuit is open no current is flowing circuit hence potential will be 120Answer: (d)
Q.28
The length of the wire is doubled, its conductance will [AFMC 2007]
0%
a) remain unchanged
0%
b) be halved
0%
c) be doubled
0%
d) be quadrupled
Explanation
if length is doubled (note wire is not stretched ), Resistance of wire will becomes two times. Conductance is reciprocal of resistance . Conductance will be halfAnswer: (b)
Q.29
The physical quantity having the dimensions [M-1 L-3T3 A2 is [AFMC 2008]
0%
a) resistance
0%
b) resistivity
0%
c) electrical conductivity
0%
d) electromotive force
Explanation
Resistance R=V/I=(J/C)/A=J/AC Dimensional formula of J ( work)=[M1L2T-2]Dimensional formula of A (current=[A]Dimensional formula of C (charge)=[AT]By substitution we get R=[M1L2T-3A-2]Resistivity=R x length=[M1L3T-3A-2]Conductivity=1/R=[M-1 L-3T3 A2]ORDimensional formula e is [AT]Dimensional formula τ is [T]Dimensional formula of n (number of electron per unit volume=[ L-3]Dimension formula of m is [M]Answer : (c)
Q.30
A battery of emf 10 V and internal resistance 3Ω is connected to a resistor. The current in the circuit is 0.5 A. The terminal voltage of the battery when the circuit is closed is [AFMC 2008]
0%
a) 10V
0%
b) 0V
0%
c) 1.5V
0%
d) 8.5V
Explanation
Voltage across the cell is called terminal voltage given by formula V=E – Ir Here E is emf r is internal resistance V=10 –(0.5)3V=10-1.5V=8.5VAnswer : (d)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page