Test: Basic & Electric Circuits- 2
20 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: Basic & Electric Circuits- 2
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A current of 3 A flows through a resistor of 20 ohms. The energy dissipated in the resistor per minute is
Power dissipated by the resistor is
P = I2 R = 32 x 20 = 180 W
Now, energy dissipated by the resistor in one hour (or 60 min)
= [180 x 60] watt-min
∴ Energy dissipated by the resistor in one minutes
A long uniform coil of a inductance L henries and associated resistance R ohms is physically cut into two exact halves which are then rewound in parallel. The resistance and inductance of the combination are
We know that,
and 
when coil is cut into two equal halves,
(Area of cross - section = constant)
So, 
Also, 
So, 
Hence, new values of resistance and inductance which are reconnected in parallel is
and 
In the circuit shown, the power dissipated in the resistor R is 1 W when only source ‘1’ is present and ‘2’ is replaced by short circuit. The power dissipated in the same resistor R is 4 W when only source '2’ is present and '1' is replaced by a short circuit When both the sources ‘1’ and ‘2’ are present, the power dissipated in R will be
We have; 
or, 
or, 
When both the sources are present, net current through R will be
I = (I2 - I1)
[as polarity of |V1| is reverse]
So, power loss in R is
If v, w, q stand for voltage, energy and charge, then v can be expressed as:
In the figure shown below, φ = power-factor angle, W = watts, VA = volt ampere and VAr = volt-ampere .reactive for an ac circuit. The correct figure is
A circuit possesses resistance R and inductive reactance XL in series, its susceptance is given by
Susceptance is the imaginary part of admittance,
or, 
Here, G = Conductance
and S = Susceptance
If V = a + jb and I = c + jd, then the power is given by
P = VI* = (a + jb).(c + jd)*
= (a + jb) (c - jd)
= (ac + bd) + j(bc - ad)
= P + jQ
Here, P = Active power
and Q = Reactive power
A 230 V, 100 W bulb has resistance RA and a 230 V, 200 W bulb has resistance RB. Here,
1. RA > RB
2. RB > RA
3. RA = 2 RB
4. RB = 2 RA
5. RA = 4 RB
From these, the correct answer is
Power, 
Since voltage ratings of both bulbs are same, therefore
PR = constant
or, PA RA = PB RB
or, 100 x RA = 200 x RB
or, RA = 2 RB
Thus, RA > RB
The voltage phapor of a circuit is 10∠15° V and the current phasor is 2∠- 45° A. The active and the reactive powers in the circuit are
S = VI* = (10∠15°) (2∠ - 45°)*
= 20∠15 + 45° = 20∠60°
= 20 (cos 60° + j sin 60°)
= (10 + j10√3) = (10 + j17.32)
= P+ jQ
In the circuit shown below, if the power consumed by the 5 Ω resistor is 10 W, then power factor or the circuit will be
Given, power consumed is
P = I2 Req
or, 
Also, 
or, |Z| = 25Ω
or, 
or, 
Hence, p.f. of given circuit is
The minimum requirements for causing flow of current are.
Carbon resistor and semiconductors have nonlinear relationship between V and I. Hence, Ohm’s law is not applicable. Also, these are not bilateral.
For a fixed supply voltage, the current flowing through a conductor will increase when its
When l is reduced. I will be increased and vice-versa.
Two registors R1, and R2 give combined resistance of 4.5 Ω when in series and 1 Ω when in parallel. The resistances are
R1 + R2 = 4.5 ...(i)
and 
∴ (R1-R2)2 = (R1 + R2)2 - 4R1R2
or, 
...(ii)
On solving equations (i) and (ii), we get
R1 = 3 Ω and R2 = 1.5 Ω
or, R1 = 1.5 Ω and R2 = 3Ω.
Which of the following is not equivalent to watts?
Two heaters, rated at 1000 W, 250 V each are connected in series across a 250 V, 50 Hz ac mains. The total power drawn from the supply would be
For series connection,
Req = R1 + R2
or, 
or, 
Given,
P1 = P2 = 1000 W
∴ Peq = 500 Watt
A 100 watt light bulb burns on an average of 10 hours a day for one week. The weekly consumption of energy will be
Energy consumption per week is
W= 100 x 10 x 7
= 7000 kW - hr
= 7 units
Assertion (A): The direction of flow of conventional current is taken opposite to that of electrons.
Reason (R): Electrons have negative charge.
Match List-I (Materials) with List-lI (Range of resistivity) and select the correct answer using the codes given below the lists:
List-I
A. Conducting materials
B. Semiconductor materials
C. insulating material
List-II
1. 100 to 102 Ω-m
2. 10-8 to 10-6 Ω-m
3. 1012to 1018 Ω-m
4. 1020 to 1030 Ω-m
Codes:
A constant current source supplies a current of 200 mA to a load of 2 kΩ When the load is changed to 100 Ω, the load current is
For a constant C.S., current will remain constant for all values of loads.
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