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Quiz 9
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Q.1
In the following figure, the reading of an ideal voltmeter V is zero. Then the relation between R, r1 and r2 is
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a) r=r2 - r1
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b) R=r1 - r2
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c) R=r1 + r2
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d) R=r1r2 / (r1 + r2 )
Explanation
Current I=2E / ( r1 + r2R) V1=E - Ir1=0 ( given)Or E=Ir1 E=2Er1 /(r1 + r2R) r1 + r2R=2r1 R=r1 -r2 Answer: (b)
Q.2
A primary cell has emf 2 volts. When short circuited, it gives current of 4 amp. Its internal resistance in ohms will be
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a)0.5
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b) 2
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c)5
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d)8
Explanation
Answer: (a)
Q.3
A cell of e.m.f E volts and internal resistance r ohm is supplying a current of i ampere to external circuit, then the terminal potential difference is
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a) E
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b) E - ir
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c)E + ir
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d)E ± ir
Explanation
Answer: (b)
Q.4
In the following figure, the value of resistor to be connected between C and D so that the resistance of the entire circuit between A and B does not change with the number of elementary sets used is( each resistance in circuit is of value R :
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a)R
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b) R(√3 -1)
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c)3R
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d)R(√3 + 1)
Explanation
the ladder is infinite. So the effective circuit is Answer:(b)
Q.5
Twelve equal resistors, each of resistance r, are connected to form a skeleton cube. Then the equivalent resistance taken between tow diagonally opposite corners is
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a) r
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b) 12r
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c)5r/6
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d)(7/12)r
Explanation
Follow any path between A and D let R be the equivalent resistance Then 6IR=5Ir R=(5/6)rAnswer: (c)
Q.6
In the above question, the equivalent resistance between the adjacent corners of any one face of the cube is
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a) r
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b) 12r
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c)5r/6
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d)(7/12)r
Explanation
Applying kirchoff's II law to mesh ABCDA zr + 2zr +zr - (x-z)r=0 x=5z Again applying kirchoff's II law to mesh A'B'C'D'A' xr + (x-z)r + xr - yr=0 3x - z=y 3(5z) - z=y y=14z VA'D'=(2x+y) rA'D'=yr rA'D'=yr / (2x+ y) y=14z and and x=5z from above rA'D'=14z / (10z+14z )=14zr/ 24z=7r/12 Answer:(d)
Q.7
In the following figure the resistance of galvanometer G is 50Ω. Of the following alternatives in which case are the currents arranged strictly in the order of decreasing magnitudes with the larger coming earlier
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a) i, i2, ig, i1
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b) i , ig, i1, i2
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c) i, i2, i1, ig
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d) ig, i1, i2, i
Explanation
Bridge is balanced ig=0. Answer: (c)
Q.8
In Wheatstone's bridge, P=9 ohms, Q=11 ohms, R=4 ohms and S=6 ohms. How much resistance must be put in parallel to the resistance S to balance bridge?
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a)24 ohms
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b) (44/9) ohms
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c)26.4 ohms
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d)18.7 ohms
Explanation
P/Q=R/SS=RQ/P by substituting values except S we get S=44/9 But actual is 6 let x be the resistance connected in parallel 9/44 \\ 1/6 + 1/x X=26.4 ohmsAnswer: (c)
Q.9
In figure to make the bulb of rating 5W, 2V glow at normal intensity, the emf E is its internal resistance is 0.46Ω
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a) 2 V
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b) 2.76 V
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c)3.75 V
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d)5 V
Explanation
Rating of bulb is 5 W and 2 Vresistance of bulb=r=V2/ P=4/5=0.8 ΩAnd current i=V/r=2/0.8=2.5A Thus current of 2.5 A should flow through bulb Let I current flow through 1.6 ohm resistance i/I=R/r 2.5/I=1.6/0.8 I=1.25A Thus battery should provide total current of 2.5+1.25=3.75 A Total resistance of the circuit R=[(0.8)(1.6) / (0.8+1.6)]+ 0.46=1 Potential of battery=Current × resistance E=3.75 × 1=3.75 V Answer: (c)
Q.10
A battery of emf having internal resistance 1Ω is connected to a external resistor of resistance 4Ω . the rate of energy dissipation in battery is
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a) 2W
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b) 4W
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c)6W
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d)8W
Explanation
total resistance in circuit=4+1=5 Ω current through battery=V/R=10/5=2 APower dissipation in battery=I2×r=4×1=4W Answer:(b)
Q.11
A current of 2A flows in a system of conductors as shown in the following figure. The potential difference VA - VB will be ( in volts)
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a) +2
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b) +1
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c) -1
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d) -2
Explanation
resistance between DAC=DBC=5 Ω Current will get equally divided in two branches=1A VD - VA=2×1=2VVD - VB=3×1=3VVD - VB -(VD - VA)=3-2VA - VB=1 VAnswer: (b)
Q.12
When a current is divided between two resistors according to Kirchhoff's law, then the heat produced is
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a)zero
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b) negligible
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c)minimum
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d)maximum
Explanation
Answer: (c)
Q.13
An electric cell is
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a) a source of charge
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b) a source of energy
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c)an energy converter
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d)both a source of charge and of energy
Explanation
Answer: (c)
Q.14
10 cells each of emf 1V and internal resistance 0.1Ω are to send maximum current through an external resistance of 100Ω, the cell be connected in
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a) series
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b) parallel
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c)mixed grouping
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d)can not say
Explanation
Answer:(a)
Q.15
The potential difference V across a filament lamp is related to the current I by V=2I + 8IThe lamp is connected in one arm of a wheatstone bridge and a resistance of 4Ω each is connected in other arms. The potential difference that must be applied to the bridge, to obtain balance is
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a) 1 V
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b) 2 V
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c) 4 V
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d) 8 V
Explanation
Let I be the current through the battery when bridge is balanced Since all other resistance are same current will equally divided in two branches current through each branch is I/2 from given equation for potential across bulb V=2(I/2) + 8(I/2)2 V=I +2I2 Resistance of bulb must be 4Ω to balance bridge Potential across bulb V=IR=(I/2)×4=2I Volts Thus 2 I=I + 2I2 I=(1/2) A The net resistance of the circuit will be 4 Ω V'=(1/2)×4=2 V Answer: (b)
Q.16
To get maximum current in a resistance of 3Ω, one can use n rows connected in parallel. Each row contains m cells connected in series. If the total number of cells is 24 and the internal resistance of the cell is 0.5Ω, then
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a) m = 12, n = 2
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b) m = 8, n=3
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c) m = 2, n = 12
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d) m = 6, n =4
Explanation
Current will be maximum if the internal resistance = external resistance Total internal resistance = mr/n = 3 m/n = 3/0.5 = 6 m = 6n Also mn = 24 6m2 = 24 n = 2 and m = 12 Answer: (a)
Q.17
A star ( as shown in figure) is made of uniform wire. The resistance of the arm EL is r ohm. The resistance of the star between the terminals C and F is
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a) 1.946r
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b) 0.937r
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c) 0.62r
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d) 3.892r
Explanation
∠ BOC = 72°, therefore∠BEC = 36° ∠FEM = 18° Side FL = 2 ×FM = 2(rsin18 ) = 0.56r Consider a ΔFEL resistance of FE and LE are in series = 2r and resistance of FL is parallel to above combination Thus total resistance of ΔFEL = (2r)×'(0.56r) / 2.5r = 0.4375r Now, the resistance of each Δ KDL, ADF and BGH same as ΔFEL Hence arrangement becomes as shown in figure There is no current through HK as potential at H and K is same So effective resistance between C and F is Answer: (b)
Q.18
Five identical resistance each of 1100Ω are connected to 220 volt as shown in the following figure. The reading of an ideal ammeter A is
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a) 1 A
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b) (3/5) A
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c) (1/5) A
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d) (2/5) A
Explanation
Current through each resistor I = 220/1100= (1/5) A Ammeter carries current only due to last three lamps. Therefore the current in ammeter is (3/5)A. Answer:(b)
Q.19
In figure the current through resistor R is
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a) 3.0 A
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b) 13.0 A
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c) 6.5 A
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d) 9.0 A
Explanation
Use Kirchoff's first law for current through R (8-5)A = 3A Answer: (a)
Q.20
A cell of e.m.f E and internal resistance r has an external resistance R connected across it. IF the terminal potential difference across the terminals of the cell is V, then
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a)r=R
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b) r=R(E/V)
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c)r=R(E-V)/V
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d)r=RV/E
Explanation
TErminal voltage V=E - Ir r=(E-V)/ IBut I=V/RAnswer: (c)
Q.21
A capacitor of 10µF ha a potential difference of 40 V across it. If it is discharged in 0.2s, the average current during discharge is
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a) 2 mA
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b) 4 mA
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c)1 mA
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d)0.5 mA
Explanation
Use i=Q/t but Q=CVi=CV/tAnswer: (a)
Q.22
An external resistance R is connected to a cell of emf E and internal resistance r. Maximum power delivered to R is
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a) E2 / r
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b) E2 / 2r
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c) E2 / 4r
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d) E2 / 8r
Explanation
power delivered P=E2r / (r+R)2 condition for maximum power is R=r Answer:(c)
Q.23
An external resistance R is connected to a cell of emf E and internal resistance r., the terminal potential difference is zero, when
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a) R > r
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b) R=r
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c) R=0
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d) R=∞
Explanation
Terminal potential=IR IR=E - ir IR=0 only if R=0 Answer: (c)
Q.24
The resistance of the circuit between A and B is ( each resistance is o r
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a)r
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b) 0.5r
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c)2r
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d)3r
Explanation
Answer: (b)
Q.25
A battery of 20 cells is charged by 220V with a charging current of 15 ampere. If the e.m.f of each cell is 2 V and internal resistance is 0.1 Ω, then the series resistance required to be placed in the circuit is
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a) 12 Ω
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b) 14 Ω
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c)10 Ω
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d)16 Ω
Explanation
charging current I=(emf of charger - emf of battery) / Total resistance 15=(220 -20×2) / ( 0.1×20 + R) 30+15R=180 R=10 ΩAnswer: (c)
Q.26
In above problem amount of power wasted as heat is
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a) 450 watt
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b) 3300 watts
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c)2700 watt
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d)2250 watt
Explanation
Power wasted P=I2 r + I2 RP=152× 2 + 152× 10P=2700 watt Answer:(c)
Q.27
The resistance of the circuit between the points A and B is
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a) R
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b) 8R/3
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c) 4R/3
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d) cannot be calculated
Explanation
Breaking the symmetry at the centre solve further Answer: (c)
Q.28
The length of a potentiometer wire required to balance the e.m.f of cell A is 60 cm. The length of the wire under the same conditions required to balance the e.m.f. if cll B is 45 Cm. If the e.m.f. of the cell B is 3.0 V, the e.m.f. of the cell A is
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a)2.25 V
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b) 2.5 V
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c)4.0 V
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d)4.5 V
Explanation
E1 / E2=l1 / l2 Answer: (c)
Q.29
Five cells, each of e.m.f E and internal resistance r are connected in series. If due to oversight, one cell is connected wrongly, then the equivalent e.m.f. and internal resistance of combination is
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a) 5E and 5r
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b) 3E and 3r
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c)3E and 5r
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d)5E and 3r
Explanation
Answer: (c)
Q.30
In figure , AB is a potentiometer wire of length 10m and resistance 2Ω. With key k open the balancing length is 5.5 m. However on closing key k the balancing length reduces to 5 m. The internal resistance of the cell E1 is :
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a)0.01 Ω
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b) 0.1 Ω
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c)0.2 Ω
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d)1.0Ω
Explanation
When the key k is open, we read the emf of the cell i.e. E1=ρ×l=ρ×5.5 When the key is closed V1=ρ×5.0 Internal resistance r=R[ (l1/l1) - 1] r=1[(E/V) - 1] r=1[(5.5ρ/5ρ) - 1]=0.1Ω Answer:(b)
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