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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 = I^{2} R = 3^{2} x 20 = 180 W
Now, energy dissipated by the resistor in one hour (or 60 min)
= [180 x 60] wattmin
∴ 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 = (I_{2 } I_{1})
[as polarity of V_{1} 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 = powerfactor angle, W = watts, VA = volt ampere and VAr = voltampere .reactive for an ac circuit. The correct figure is
A circuit possesses resistance R and inductive reactance X_{L} 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 R_{A} and a 230 V, 200 W bulb has resistance R_{B}. Here,
1. R_{A} > R_{B}
2. R_{B} > R_{A}
3. R_{A} = 2 R_{B}
4. R_{B} = 2 R_{A}
5. R_{A} = 4 R_{B}
From these, the correct answer is
Power,
Since voltage ratings of both bulbs are same, therefore
PR = constant
or, P_{A} R_{A} = P_{B} R_{B}
or, 100 x R_{A} = 200 x R_{B}
or, R_{A} = 2 R_{B}
Thus, R_{A} > R_{B}
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 = I^{2 }R_{eq}
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 viceversa.
Two registors R_{1}, and R_{2} give combined resistance of 4.5 Ω when in series and 1 Ω when in parallel. The resistances are
R_{1 }+ R_{2} = 4.5 ...(i)
and
∴ (R_{1}R_{2})^{2} = (R_{1} + R_{2})^{2 } 4R_{1}R_{2}
or, ...(ii)
On solving equations (i) and (ii), we get
R_{1} = 3 Ω and R_{2} = 1.5 Ω
or, R_{1} = 1.5 Ω and R_{2 }= 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,
R_{eq} = R_{1} + R_{2}
or,
or,
Given,
P_{1} = P_{2 }= 1000 W
∴ P_{eq }= 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 ListI (Materials) with ListlI (Range of resistivity) and select the correct answer using the codes given below the lists:
ListI
A. Conducting materials
B. Semiconductor materials
C. insulating material
ListII
1. 10^{0} to 10^{2} Ωm
2. 10^{8} to 10^{6} Ωm
3. 10^{12}to 10^{18 }Ωm
4. 10^{20} to 10^{30} Ωm
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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