Test: Basic & Electric Circuits- 2


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Test: Basic & Electric Circuits- 2 - Question 1

A current of 3 A flows through a resistor of 20 ohms. The energy dissipated in the resistor per minute is

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 1

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

Test: Basic & Electric Circuits- 2 - Question 2

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 2

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 

Test: Basic & Electric Circuits- 2 - Question 3

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 3

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



 

Test: Basic & Electric Circuits- 2 - Question 4

If v, w, q stand for voltage, energy and charge, then v can be expressed as:

Test: Basic & Electric Circuits- 2 - Question 5

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 5

Test: Basic & Electric Circuits- 2 - Question 6

A circuit possesses resistance R and inductive reactance XL in series, its susceptance is given by

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 6

Susceptance is the imaginary part of admittance,


or, 
Here, G = Conductance

and S = Susceptance

Test: Basic & Electric Circuits- 2 - Question 7

If V = a + jb and I = c + jd, then the power is given by

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 7

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

Test: Basic & Electric Circuits- 2 - Question 8

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 8

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

Test: Basic & Electric Circuits- 2 - Question 9

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 9

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

Test: Basic & Electric Circuits- 2 - Question 10

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 10

Given, power consumed is
P = IReq
or, 
Also, 
or, |Z| = 25Ω
or, 
or, 
Hence, p.f. of given circuit is

Test: Basic & Electric Circuits- 2 - Question 11

The minimum requirements for causing flow of current are.

Test: Basic & Electric Circuits- 2 - Question 12

Ohm’s law is applicable to

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 12

Carbon resistor and semiconductors have nonlinear relationship between V and I. Hence, Ohm’s law is not applicable. Also, these are not bilateral.

Test: Basic & Electric Circuits- 2 - Question 13

For a fixed supply voltage, the current flowing through a conductor will increase when its

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 13


When l is reduced. I will be increased and vice-versa.

Test: Basic & Electric Circuits- 2 - Question 14

Two registors R1, and R2 give combined resistance of 4.5 Ω when in series and 1 Ω when in parallel. The resistances are

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 14

R1 + R2 = 4.5 ...(i)
and 
∴ (R1-R2)2 = (R1 + R2)- 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.

Test: Basic & Electric Circuits- 2 - Question 15

Which of the following is not equivalent to watts?

Test: Basic & Electric Circuits- 2 - Question 16

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 16

For series connection,
Req = R1 + R2
or, 
or, 
Given,
P1 = P= 1000 W
∴ Peq = 500 Watt

Test: Basic & Electric Circuits- 2 - Question 17

A 100 watt light bulb burns on an average of 10 hours a day for one week. The weekly consumption of energy will be

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 17

Energy consumption per week is
W= 100 x 10 x 7
= 7000 kW - hr
= 7 units

Test: Basic & Electric Circuits- 2 - Question 18

Assertion (A): The direction of flow of conventional current is taken opposite to that of electrons.
Reason (R): Electrons have negative charge.

Test: Basic & Electric Circuits- 2 - Question 19

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:

Test: Basic & Electric Circuits- 2 - Question 20

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

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 20

For a constant C.S., current will remain constant for all values of loads.

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