MCQs On Direct Current (DC) Electrical Circuit

28
MARCH, 2018

1) A DC circuit has ___________ as a load.

1. Resistance
2. Inductance
3. Capacitance
4. All of the above
EXPLANATION
DC Circuit has no field changing effect. So, there will be no inductance and capacitance effect in the circuit.

2) The purpose of load in an electric circuit is to

1. Increase the circuit current
2. Utilize electrical energy
3. Decrease the circuit current
4. None of the above
EXPLANATION
An electrical load is an electrical component or portion of a circuit that consumes electric power.

3) Electrical appliances are not connected in series because

1. Series circuit is complicated
2. Power loss is more
3. Appliances have different current ratings
4. None of the above
EXPLANATION
Electrical appliances are not connected in series because current will be same through all the appliances connected whereas appliances have different current ratings and also that voltage decreases as more no. Of appliances are connected.

4) Electrical appliances are connected in parallel because it

1. Is a simple circuit
2. Draws less current
3. Results in reduced power loss
4. Makes the operation of appliances independent of each other
EXPLANATION
When appliances are connected in a parallel arrangement, each of them can be put on and off independently.

5) Three resistance 14.5 Ω, 25.5 Ω and 60 Ω are connected in series across 200 V. What will be the voltage drop across 14.5 Ω

1. 29 V
2. 13.5 V
3. 14 V
4. 18 V
EXPLANATION
In this circuit, total current is calculated as
R = 14.5+25.5+60
R = 100 Ω
I = V/R
I = 200/100
I = 2 A
Now, voltage drop across 14.5 Ω resistance is
V = IR
V = 2×14.5
V = 29 V

6) Two bulbs B1 100 W, 200 V and B2 40 W, 200 V are connected in series across 200 V battery, the total circuit resistance will be

1. 1000 Ω
2. 400 Ω
3. 1400 Ω
4. 135 Ω
EXPLANATION
In this circuit,
For B1
P1 = V2/R
R1 = V2/P
R1 = (200)2 / 100
R1 = 400 Ω
For B2
P2 = V2/R
R2 = (200)2 / 40
R2 = 1000 Ω
Total resistance in circuit R = R1 + R2
R = 1400 Ω

7) When the number of resistance are connected in parallel, the total resistance is

1. Greater than the smallest resistance
2. Between the smallest and greatest resistance
3. Less than the smallest resistance
4. None of the above
EXPLANATION
If two resistances in parallel are equal and of the same value, then the total or equivalent resistance is equal to half the value of one resistor.

8)  When cells are arranged in parallel

1. The current capacity increases
2. The current capacity decreases
3. The e.m.f decreases
4. The e.m.f increases
EXPLANATION
In a parallel circuit there is more than one path for the current to flow along. When cells are connected in parallel the total value of the cells is the same as each individual cell. For two 1.5 V cells in parallel, the voltage stays at 1.5 V, but the life (current capacity) of the battery is doubled.

9) A cell of e.m.f E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The instant resistance of the cell is

1. 2(E – V)V/r
2. 2(E – V)r/E
3. (E – V)r
4. ((E – V)/V)r
EXPLANATION
e=I(R+r)
V=IR
∴ eV=R+VR
=1+rR
=rR
=(ev−1)
∴ r=R(eV−1)

10) An external resistance R is connected to a cell of internal resistance r. The maximum current flows in the external resistance when

1. R < r
2. R > r
3. R = r
4. None of the above
EXPLANATION
The external resistance (R) is connected with the internal resistance (r) of the cell in series connection. So, the total resistance is ( R + r ). When the total resistance is low, then there is a maximum current flow in that circuit. If we apply the option (A) and (B), then the overall resistance increases than the option (C). Hence, if the external resistance (R) and the internal resistance (r) are equal, then the maximum current flows through the circuit.

11) A wire has a resistance of 12 Ω. It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is

1. 6 Ω
2. 3 Ω
3. 12 Ω
4. 24 Ω
EXPLANATION
Here, wire is connected by bending it in circle. After then, Taking two points at any diameter means circle is divided in two equal half that makes it parallel connected. So, 12 Ω is divided into two equal i.e. 6 Ω resistance each. By solving these two parallel connected resistance we get 3 Ω as equivalent resistance.

12) The total resistance of circuit by connecting 50 resistance of 1/4 Ω in parallel is

1. 50/4 Ω
2. 4/50 Ω
3. 200 Ω
4. 1/200 Ω

EXPLANATION
1/4=0.25 ohm
1/R=1/0.25+1/0.25+••••••••••50 resisters
1/R=1/0.25*50=200 ohm
R= 1/200 ohm

13) The internal resistance of a cell of e.m.f 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell is

1. 0.5 V
2. 1.95 V
3. 1.9 V
4. 2 V
EXPLANATION
Here Current I = (E / (r+R) )
I = 2 / (3.9+0.1)
I = 2 / 4
I = 0.5

14) Three 2 Ω resistors are connected to form a triangle. The resistance between any two corners is

1. 6 Ω
2. 2 Ω
3. (3/4) Ω
4. (4/3) Ω
EXPLANATION
The total resistance of the triangle and between any two corners will be 1.33 ohm.
2 || (2+2)
2 || 4
(2×4)/(2+4)
8/6 = 1.33333

15) A cell with zero internal resistance and e.m.f of 2 V is connected across 2, 3 and 5 Ω. The potential difference across 3 Ω resistor will be

1. (2/3) V
2. 0.6 V
3. 3 V
4. 6 V
EXPLANATION
First find the total current within it.
Rt = R1 + R2 + R3
Rt = 10 Ω
V = IRt
I = V / R
I = 2/10
I = 0.2 A
Now, voltage drop across 3 ohm resistor.
V = IR2
V = 0.2×3
V = 0.6 V

16) Two bulbs 200 W and 100 W are connected across 220 V supply in series. Find the power consumed by both bulbs

1. 300 W
2. 200 W
3. 66 W
4. 50 W
EXPLANATION
Here first calculate the resistance of the circuit. i.e.
Rt = R1 + R2
R1 = (V)2 / P1
R1 = (220)2 / 200
R1 = 242
R2 = (V)2 / P2
R2 = (220)2 / 100
R2 = 484
Rt = 242 + 484
Rt = 726

Now, Calculate the total Power
Pt = (V)2 / Rt
Pt = (220)2 / 726
Pt =66 W

17) An electrical fan and heater are marked as  100 W and 1000 W respectively at 220 V supply. What will be the resistance of heater

1. Zero
2. Greater than that of the fan
3. Less than that of the fan
4. Equal to that of the fan
EXPLANATION
For constant voltage, we know that
P ∝ 1/R
So higher the power, lower will be the resistance.

18) If the internal resistance of the cell connected in circuit is greater than the external circuit, higher current will be obtained by arranging cell in ______

1. Series
2. Parallel
3. Series-Parallel
4. Any connection
EXPLANATION
As per the above statement, the internal resistance of the cell is larger than the external resistance. So, if we connect the cells in parallel, then the overall internal resistance decreases. We know that, in low resistance path the current flow is more. Hence, we need to group the cells in parallel to make the current flow as high.

19) The cell e.m.f depends upon

1. Internal resistance
2. External resistance
3. Electrolyte and electrode of the cell
4. All of the above
EXPLANATION
Emf of a cell is the energy required or the work done to drive a unit charge to pass through a given circuit.It depends on
Nature of the material of electrode.
Internal resistance of the cell.
Distance between the electrodes.

20) How many combination scheme of three resistors can be created

1. 2
2. 3
3. 4
4. 5
EXPLANATION
Three resistors yield four combinations.

21) Two resistance in parallel gives combine effect of 6/5 Ω. If one resistance broke, the effective resistance becomes 2 Ω. The broken resistance will be

1. 3/5 Ω
2. 2 Ω
3. 3 Ω
4. 6 Ω
EXPLANATION
Apply series & parallel combination to calculate the value of resistance.

22) Two identical resistors connected in series send 10 A through 5 Ω resistor. When same cell connected in parallel sends 8 A. The internal resistance of each cell will be

1. 0 Ω
2. 5 Ω
3. 2.5 Ω
4. 7.5 Ω
EXPLANATION
When two identical cells are connected in parallel.
2E = (5 + 2r) × 10
When these cells are connected in series.
E = (5 + r/2) × 8
Rearrange both equation and calculate value of r
We get r = 2.5 Ω

23) Two batteries of same e.m.f and different internal resistance are connected in series with an external resistance and current is 3 A. When polarity of one is reversed, the current is 1 A. The ratio of e.m.f is

1. 2.5
2. 2
3. 3
4. 6
EXPLANATION
Here E1 and E2 are equal. So when connected in same polarity.
(E1+E2)/R+r = 3 A
Now, when polarity is reversed.
(E1-E2)/R+r = 1 A
Rearranging equations
(E1+E2)/(E1-E2) = 3/1
E1+E2 = 3(E1-E2)
E1/E2 = 2

24) When n resistance of r ohm connected in parallel give R total resistance. If all those resistance are connected in series, total resistance will be

1. n × R
2. R / n2
3. n2× R
4. R / n
EXPLANATION
When connected in parallel.
R = r / n i.e. r = n × R
When resistance of each resistance is r×R in series and there are n total number of resistance. Total resistance will be
n×n×R = n2 × R

25) Kirchhoff’s current law at a junction deals with

1. Conversation of energy
2. Conversation of momentum
3. Conversation of charge
4. Conversation of angular momentum
EXPLANATION
No answer description required for this question.

26) Kirchhoff’s voltage law deals with

1. Conversation of energy
2. Conversation of momentum
3. Conversation of charge
4. Conversation of angular momentum
EXPLANATION
No answer description required for this question.

27) What will be the internal resistance of cell, if it supplies 0.9 A current through 2 Ω resistor and 0.3 A current through 7 Ω resistor.

1. 0.9 Ω
2. 1.2 Ω
3. 1 Ω
4. 0.5 Ω
EXPLANATION
Let E and r be the e.m.f and internal resistor of cell.
For first case,
0.9 = (E/2+r)
For second case,
0.3= (E/7+r)
Solving equations, we get
r = 5 Ω

28) A battery of 50 V connected across 10 Ω resistor draws 4.5 A current. The internal resistance of the battery is

1. 0 Ω
2. 5 Ω
3. 0.5 Ω
4. 4.05 Ω
EXPLANATION
E = v + iR
= 50 + 4.50 x 10
= 95
internal resistance r = R ( E / v – 1 )
= 4.5 ( 95 / 50 )-1
= 4.05

29) Four cells, each of having internal resistance 1 Ω are connected in parallel. The battery resistance will be

1. 4 Ω
2. 0.25 Ω
3. 2 Ω
4. 1 Ω
EXPLANATION
The parallel combination of resistors with make resistance equal to 0.25 Ω

30) Two voltmeters V1 and V2 are connected in series across DC source. The V1 reads 80 V and has per volt resistance of 200 Ω while V2 has a total resistance of 32 kΩ. The line voltage is

1. 120 V
2. 160 V
3. 220 V
4. 240 V
EXPLANATION
Resistance of voltmeter V1, R1 = 80 × 200 = 16000
Circuit current I = V1/R1 = 80/16000 = 5 mA
Reading of V2 = I×R1 = 5 mA × 32 kΩ = 160 V
Line voltage = V1+V2 = 160+80 = 240 V

31) Resistance 1 Ω, 2 Ω, and 3 Ω are connected in the form of triangle. If a cell of e.m.f 1.5 V and negligible internal resistance is connected across 3 Ω, the current through this resistance will be

1. 0.25 A
2. 0.5 A
3. 1.5 A
4. 2.5 A
EXPLANATION
Solve it by yourself considering the conditions given above.

32) what will be the current in 2 seconds from a wire of having 10 mC charge flow and 2.5 mm cross sectional area.

1. 25 mA
2. 5 mA
3. 20 mA
4. 4 mA

EXPLANATION
No answer description required for this question.

33) When two cells whether connected in series or parallel have same current through an external resistance of 2 Ω.What will be their internal resistance.

1. 0.5 Ω
2. 1 Ω
3. 1.5 Ω
4. 2 Ω
EXPLANATION
Let E and r be the e.m.f and internal resistance of cell respectively. Then
2E/(2+2r) = E/(2+r/2)
or

2/(2+2r) = 1/(2+r/2)
On solving we get r = 2 Ω

34) Current divides itself through resistance connected in parallel by

1. Inverse ratio of resistance
2. Direct ratio of resistance
3. Direct ratio of potential
4. None of the above
EXPLANATION
The resistive value of each branch determines the amount of current flowing within that branch. Greater the resistance, smaller will be the current flowing through it.

35) If a DC supply delivers maximum power to external resistance. The ratio of its internal and external resistance r/R will be

1. 1 : 1
2. 1 : 2
3. 2 : 1
4. 1 : 1/2
EXPLANATION
No answer description required for this question.

36) Two bulb rated at 25 W, 110 V and 100 W, 110 V are connected across 220 V supply. What will happen ?

1. 100 W bulb will burn out
2. 25 W bulb will burn out
3. Both bulb will burn out
4. No bulb will burn out
EXPLANATION
Using the formula W = E^2/R, we can work out that the resistance of the 25W lamp is 484 ohms, and that of the 100W lamp is 121 ohms. From this we can work out that at 110V, the 25W bulb draws 22mA and the 100W bulb draws 91mA.
If you put the two lamps in series across 220V, the total resistance is 605 ohms, and so a current of 36.36mA flows through both lamps. The voltage drop across the 25W bulb is therefore now 176 volts. It is overloaded and will burn out.

37) Three equal resistors when connected in series dissipates 10 W power, how much power will it dissipates when the same resistors are connected across same potential in parallel.

1. 90 W
2. 30 W
3. 180 W
4. 27 W
EXPLANATION
By changing from series to parallel, the total resistance decreases by a factor of 9. So by applying the power equation P = (V2)/R, and assuming a constant supply voltage, the power dissipated in the resistor will increase by a factor of 9. So the new dissipation is 90W.

38) In a circuit two cells of 1.5 V and 2 V e.m.f having internal resistance of 2 Ω respectively are connected in parallel so as to send current in the same direction through an external resistance of 5 Ω. The current will be

1. 17/5 A
2. 12/5 A
3. 5/17 A
4. 5/12 A
EXPLANATION
By current divider rule through 5 resistor will be I2 = (2/3) / ((2/3) +5) I2 = 5 / 17 A

39) A 40 W bulb is connected in parallel with room heater. What will happen if we turn off bulb ?

1. Heater output will increase
2. Heater output will decrease
3. Heater output will remains same
4. None of the above
EXPLANATION
Heater output will remain same, as in parallel combination current flow through appliance depends upon its own resistance.

40) The total conductance of 0.1 S conductance in series is

1. 1 S
2. 0.1 S
3. 0.05 S
4. 0.2 S
EXPLANATION
G = 1 / (1/G1 + 1/G2) = 1 / (1/0.1 + 1/0.1) = 1/(10 + 10) G = 0.05 S

41) hree conductance G1 , G2 , G3 are connected in parallel. If the total circuit current is 4 A, the current in G1 will be

1. 1.2 A
2. 2 A
3. 0.8 A
4. 4 A
EXPLANATION
As conductance is reciprocal of resistance, so if conductance is assumed to be connected in parallel. Series resistance formula will apply on it. Thus,
G = G1 + G2 + G3 G = 0.5+0.3+0.2 G = 1 S Current in G1 = 4 × (G1 /G2 ) Current in G1 = 4 × (0.5/0.3) Current in G1 = 2 A

42) A resistor in a circuit has a value of 560 Ω. It is desired to decrease its value to 344 Ω. The resistance to be connected with it in parallel will be of

1. 740 Ω
2. 1090 Ω
3. 892 Ω
4. 540 Ω
EXPLANATION
Rt = R×560 / R+560
OR
344 = R×560 / R+560
R = 892 Ω

43) An ammeter has a resistance of 50 Ω and requires 10 mA for full scale deflection. The shunt resistance S required to allow the meter to measure up to  100 mA is

1. 5.56 Ω
2. 2.5 Ω
3. 15.75 Ω
4. 8.75 Ω
EXPLANATION
Here in this condition, full scale voltage required for it will be:
IR = 10 mA × 50 = 0.5 V
Now when circuit resistance becomes 100 mA, the current through shunt will be
100 – 10 = 90 mA
S = 0.5 / 90 mA
S = 5.56 Ω
44) The power dissipated in a resistor can be calculated in term of conductance as

1. I2 G
2. I2 / G
3. G2 / I
4. None of the above
EXPLANATION
Conductance is reciprocal of resistance. So, its power calculation formula will also be with reciprocal of resistance. i.e. I2/G

45) Two wires of same properties with having cross sectional area of 3:1 joined together in series. The resistance of thicker wire is 3 Ω. What will be the resistance of thinner wire.

1. 5/2 Ω
2. 40/3 Ω
3. 40 Ω
4. 100 Ω
EXPLANATION
R1 = ρ×(l/3A) and R2 = ρ× (l/A) R2/R1 = 3 A / A R2 / R1 = 3 R2 = 3×R1 R2 = 3×10 = 30 Ω Total resistance = R1+R2 = 40 Ω

46) The resistance of each arm of Wheatstone bridge is 10 Ω. A resistance of 10 Ω is connected in series with the galvanometer. The equivalent resistance across the battery will be

1. 40 Ω
2. 20 Ω
3. 30 Ω
4. 10 Ω
EXPLANATION
The equivalent resistance will be 10 ohm.

47) If a galvanometer is short circuited in a balanced Wheatstone bridge, the current in various resistors of Wheatstone bridge will

1. Increased
2. Decreased
3. Not change
4. Cannot say anything
EXPLANATION
In a balanced Wheatstone bridge, there is no current through galvanometer. So, by short circuiting, there will be no effect on circuit resistance current.

48) If a galvanometer is open circuited in a balanced Wheatstone bridge, the current in various resistors of Wheatstone bridge will

1. Increased
2. Decreased
3. Not change
4. Cannot say anything
EXPLANATION
In a balanced Wheatstone bridge, there is no current through galvanometer. So, by open circuiting, there will be no effect on circuit resistance current.

49) A resistor of 5 Ω is connected in series with a parallel combination of N number of resistors, each of 5 Ω value. If the total resistance of the circuit is 6 Ω. The number (N) of resistors in parallel will be

1. 5
2. 10
3. 15
4. 20
EXPLANATION
Let N no. of resistors are connected in parallel. The equivalent of parallel resistors will be 5/N
Now, total resistance will be = 5+5/N = 6 Ω(Given)
By rearranging equation, we get
N = 5
50) Two cells of same e.m.f E but of different internal resistances r1 and r2 are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is

1. r1+r2
2. r1– r2
3. r2– r1
4. r1r2
EXPLANATION
Circuit current I = 2E / (R+r1+r2) Drop across first cell = E – Ir1 = 0 or E – ( 2E / (R+r1+r2) ) × r1 = 0 Rearranging and solving it gives R = r1 – r2

51) Two cells whether joined in parallel or series gives the same current when connected to an external resistance of 1 Ω. The internal resistance of each cell will be

1. 1 Ω
2. 2 Ω
3. 4 Ω
4. 8 Ω
EXPLANATION
Let each cell have e.m.f E and internal resistance r
When cells in series Circuit current I1 = (2E / 1+2r)
When cells in parallel Circuit current I2 = (E / (1+r/2))
As, we are told that current is same that means I1 = I2
So,
(2E / 1+2r) = (E / (1+r/2))

By rearranging and solving, we get
r = 1 Ω
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