SSC JE 2010 Electrical question paper with Explained Solution
Ques 1. The value of voltage across the diode in figure given below is
Depend upon the value of R
There will be no current in the circuit as the diode is reverse biased and diode in reversed biased offer infinite resistance. You can think of a reverse-biased diode as an open circuit.
So, whole voltage will across the diode i.e 8V
Ques 2. The temperature coefficient of an intrinsic semiconductor is
Like this of Metal
Properties of intrinsic semiconductors
They have four valence electrons
Their properties lie between that of conductors and insulators.
Their conduction property can be varied by varying the temperature.
They have the negative temperature coefficient of resistance.
Their properties can be varied by adding impurities to the semiconductor.
The resistivity of a semiconductor is greater than a conductor and less than an insulator.
Ques 3. A 4-pole, 1200 rpm DC lap wound generator has 1520 conductors. If the flux per pole is 0.01 weber, the emf of generator is
EMF Of generator is given by
E = PΦZN/60A
Φ = flux per pole in Weber
P = number of poles
Z = total number of armature conductor
N = rotation speed of the armature in revolution per minute (r.p.m)
A = number of parallel paths (Number of parallel paths in lap winding, A = P)
E = (4 x 0.01 x 1520 x 1200)/(60 x 4)
= 304 V
Ques 4. In a 3-phase induction, motor starting torque will be maximum when
R2 = 1/X2
R2 = X2✔
R2 = X22
R2 = √X2
The equation of torque is
When slip s = R2 / X2, the torque will be maximum and this slip is called maximum slip Sm and it is defined as the ratio of rotor resistance to that of rotor reactance.
If supply voltage V is kept constant, then flux ɸ and E2 both remain constant. Hence,
At starting S = 1, so the maximum starting torque occurs when rotor resistance is equal to rotor reactance. So it can be proved that maximum starting torque is obtained when rotor resistance is equal to standstill rotor reactance. i.e. R22 + X22 =2R22.
Ques 5. The ratio of resistance of a 100 W, 220 V lamp to that of a 100 W, 110 V lamp will be at respective voltages
For 100 W, 220 V
R1 = V2/P = 2202/100 = 484
For 100 W, 110 V
R2 = V2/P = 1102/100 = 121
Ratio of resistance R1 and R2
R1/R2 = 484/121 = 4
Ques 6. Two sinusoidal equations are given as
E1 = A sin(ωt + π/4)
E2 = B sin(ωt – π/6)
The phase difference between the two quantities is:
Phase difference Φ = Φ1 – Φ2
= π/4 – (-π/6) = 5π/12 = 5 x 180°/12 = 75°
Ques 7. The moderator used in fast breeder reactor is
None of the above✔
A breeder reactor is a nuclear reactor that produces more fissionable material than it consumes to generate energy. This special type of reactor is designed to extend the nuclear fuel supply for electric power generation. All fast neutron reactor designs use liquid metal as the primary coolant, to transfer heat from the core to steam used to power the electricity generating turbines.
A fast reactor needs no moderator to slow down the neutrons at all, taking advantage of the fast neutrons producing a greater number of neutrons per fission than slow neutrons. For this reason, ordinary liquid water, being a moderator as well as a neutron absorber, is an undesirable primary coolant for fast reactors.
Ques 8. The ratio of the puncture voltage to the flashover voltage of an insulator is
Equal to one
Lower than One
Greater than One✔
Insulators are rated by three voltages:
Working voltage (or rated voltage)
The working voltage rating is the voltage at which an insulator is designed to bear the steady state voltage stress. If the line voltage is VLL, the working voltage will be VLL/√3.
The flashover voltage is the voltage at which flashover occurs through air surrounding the insulator or the breakdown of gaseous medium is called as flashover.
The puncture voltage is the voltage at which the insulator breaks through between conductor and pin. It destroys the insulator or Breakdown of the solid insulator is called Puncture voltage. This rating is determined by applying the voltage while insulator emerges in oil. This is done because before the puncture, there will be flashover. Flashover voltage is less than puncture voltage and higher than working voltage of insulators.
The ratio of puncture strength to flashover voltage is called the “safety factor” of the part or of the insulator against puncture. This ratio should be high enough (>1) to provide sufficient protection for the insulator from puncture by the transients.
Ques 9. Bucholtz relay cannot be used on
5000 kV transformer
1000 kV transformer
Three phase transformer
Air cooled transformer✔
The Buchholz relay is a gas operated relay used for the protection of oil immersed transformers against all the types of internal faults. It is named after its inventor, Buchholz. It is used for the protection of a transformer from the faults occurring inside the transformer, such as impulse breakdown of the insulating oil, insulation failure of turns etc.
Buchholz relay can be used only in oil immersed transformer having a conservative tank. Due to economic considerations, Buchholz relay is not provided for the transformers having below 500 KVA.
Ques 10. An ammeter is obtained by shunting a 30Ω galvanometer with 30Ω resistance. What should additional shunt be connected across it to double the range?
The value of the shunt resistance required to convert the galvanometer into an ammeter
Where Ig = the current for full-scale deflection in the galvanometer
SSC JE 2011 Electrical question paper with Explained Solution
Ques 1. If in an R-L-C series circuit the current lags the applied voltage by 60° then
XL – Xc = R/√3
XL – Xc = √3R✔
XL = Xc = R
XL – Xc = R
Since Angle between V and I is 60°
Then Power factor angle = θ = 60°
For series RLC Circuit the phase difference between the current and the voltage is
tanθ = (XL – Xc)/R ……. (Since XL > Xc)
tan60° = (XL – Xc)/R
√3R = (XL – Xc)
Ques2. A lossy capacitor with loss angle of 0.01 radian, draws a current of 0.5 A when supplied at 1000 V from a sinusoidal voltage source. The active power consumed by the capacitor is
Loss angle = 0.01 radian
θL = 0.01 x 180/π = 0.57 ———- (Converting radian into degree)
P.F. angle = 90 – 0.57 = 89.43
Active power consumed = V I cosθ
= 1000 × 0.5 × cos(89.427)
= 4.999 ≅ 5
Ques 3. An AC voltage source with an internal impedance Z 1 is connected to a load of impedance Z2. For maximum power transfer to the load, the condition is
Z2 = Z1
Z2* = Z1
Z2 = Z1*✔
For maximum power transfer, input Impedance must be equal to the conjugate of output Impedance.
As we know that The impedance of an input of something to which a signal is applied is a measure of how much power that input will tend to draw (from a given output voltage). This impedance is known as the load impedance.
Therefore Z2 = Z1*
Ques 4. The resonant frequency of the AC series circuit shown in figure given below, in Hz is
When the coils are joined in series with additive flux the equivalent inductance is
Leq = L1 + L2 – 2M = 2 + 2 – 2 x 1 = 2H
The resonant frequency of the AC series circuit in Herz is given as
F = 1/2π√LC
Where L = 2H & C = 2F
F = 1/2π√LC = 1/2π√4
Ques 5. A current wave starts at zero, rises instantaneously, then remains at a value of 20 A for 10 sec, then decreases instantaneously, remaining at a value of – 10 A for 20 sec, and then repeats this cycle. The RMS value of the wave is
RMS Value of current of sine wave is
Ques 6. Two incandescent bulbs of rating 230 V, 100 W and 230 V, 500 W are connected in parallel across the mains. As a result what will happen?
100 W bulb will glow brighter
500 W bulb will glow brighter✔
Both the bulbs will glow equally bright
Both the bulbs will glow dim
In a parallel connection, the voltage across each element is same. So when 100W bulb and 500W bulb are connected in parallel, the voltage across them will be same (230 V in the given case). To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation
P = V2/R
R100 = 2302/100 = 529 Ohm
R500 = 2302/500 = 105.8 Ohm
Since the voltage is same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 500W bulb.
Therefore 500-watt bulb will glow brighter.
Ques 7. In two wattmeter method of measurement of three- phase power of a balanced load, if both the wattmeters indicate the same reading, then the power factor of the load is
Power factor of the wattmeter is given as
Reading of Wattmeter 1 “W1”
W1 = VLIL Cos(30 + Φ)
Reading of Wattmeter 2 “W2”
W2 = VLIL Cos(30 – Φ)
When load P.F cosΦ = 1 then Φ = 0° so that
W1 = W2 = VLIL Cos30
Ques 8. Measurement of ________ is affected by the presence of thermo-emf in the measuring circuit
In the galvanometer circuit, the dissimilar metals come in contact and generate the thermal e.m.f.s. Such thermal e.m.f.s may cause the en’0rs while measuring low-value resistances. To prevent this, more sensitive galvanometers having copper coils and copper suspension systems are used.
Ques 9.The response time of an indicating instrument is determined by its
Support type of Moving system
The major function of the damping system is to produce a damping force while the moving system is in motion. The damping force should be of such a magnitude that the pointer of the moving system comes to its final steady value quickly without any oscillation.
The response time of an indicating instrument is determined by Damping system.
If the moving system reaches its final position rapidly and smoothly without oscillations, the instrument is said to be critically damped.
If the moving system oscillates about the final stead, position with a decreasing amplitude and take some time to come to rest then the instrument is said to be underdamped.
The instrument is said to be overdamped if the moving system moves slowly to its find steady position.
In practice, slightly underdamped systems are preferred.
Ques 10. The ratio of the reading of two wattmeters connected to measure active power in a balanced 3-phase load is 2:1. The power factor of the load is
SSC JE 2012 Electrical question paper with Explained Solution
Ques 1. The emf induced per phase in a three-phase star connected synchronous generator having the following data
Distribution factor = 0.955
Coil-span factor = 0.966
Frequency = 50 Hz
Flux per pole = 25 mwb
Turns per phase = 240, then emf per phase is
E.M.F equation of an alternator is given as
E = Kc Kd √2π f Φ Np
Or E = 4.44 Kc Kd f Φ Np ………….. (since √2π = 4.44)
Kc = Coil span factor
Kd = Distribution factor
Φ = Flux per pole
f = frequency
Np = Turns per phase
E = 4.44 × 0.955 × 0.966 × 50 × 25 × 10-3 × 240
E = 1228.1 volts
Ques 2. In a 1-phase transformer, the copper loss at full load is 600 watts. A half of the full load the copper loss will be
Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.
So (Pcu)full load = 600 watts
(Pcu) half load = 600/4 = 150 watts
Ques 3. An autotransformer used with sodium vapour lamp should have high
Leakage Reactance of winding✔
A sodium vapour lamp is also called as the cold-cathode low-pressure lamp which gives high luminous output about three times higher than the other lamp.
For starting discharge through the lamp, it is essential that the striking voltage should be higher than the normal working voltage of the lamp. This high voltage is taken from a high reactance transformer or auto-transformer which has poor voltage regulation. Thus when discharge in the lamp takes place, the lamp current increase due to the decrease in the resistance of the gas in the tube and output voltages of the auto-transformer fall. The lamp then continues to operate normally.
Ques 4. In an auto-transformer, the number of turns in primary winding is 210 and in secondary winding is 140. If the input current is 60 A, the current in output and in common winding are respectively.
40 A, 20A
40A, 100 A
90A, 30 A✔
Auto-transformer ratio is given as
I1/I2 = N2/N1
60/I2 = 140/210
I2 = 90A
Current in common Winding Ic
Ic = (N2 – N1)I1/N2
= 60 x (210 – 140)/140
= 30 A
Ques 5. A 3-phase transformer has its primary connected in delta and secondary in star. Secondary to primary turns ratio per phase is 6. For a primary voltage of 200 V, the secondary voltage would be
Transformer ratio is given by
N1/N2 = V1/V2
0r N2/N1 = V2/V1
6/1 = V2/200
V2 = 1200 volt per phase
For star connection line voltage is
VL = √3 x 1200 = 2078.46 V
Ques 6. A resistance and another circuit element are connected in series across a dc voltage V. The voltage and zero after time. The other element is pure
Both (a) and (c)
Why inductor behave as short circuit for DC voltage?
An inductor has a reactance equivalent XL = 2πfL.Since a DC supply will have frequency = 0, the reactance is 0 and the inductor would be short.
Imagine a circuit of three elements in series – an input voltage source, inductor and resistive load. The input source has been producing some constant (DC) voltage; If an inductor is connected with DC source, it will not act as inductor but acts like as a simple resistor. The current through it will depend on the resistance of the inductor. Now increase the input voltage to a higher value.
The inductor immediately converts the voltage change to an equivalent “antivoltage” and applies it contrary to the input voltage change, the voltage “produced” by the inductor “jumps” with a magnitude equal to the input voltage change. Thus we have two voltage sources (an “original” and “cloned”:) contrary connected in series and neutralizing each other; as a result, the total (effective) voltage and accordingly the current do not change. After that, the current will increase exponentially (1 – e-t) and reach a constant value. Current will be maximum and inductor will behave like a short circuit.
Ques 7. For RLC series resonance the current is
Minimum at leading P.F
Minimum at Lagging P.F
Maximum at unity P.F
Maximum at leading P.F✔
The total impedance of the series LCR circuit is given as
Z = R + j (X1 – X²)
where X1 is inductive reactance
and X2 is capacitive reactance.
At a particular frequency (resonant frequency), we find that X1=X2 because resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. Therefore, the phase angle between voltage and current is zero and the power factor is unity.
Thus, at the resonant frequency, the net reactance is zero because X1=X2. The circuit impedance Z becomes minimum and is equal to the resistance R. “Since the impedance is minimum, the current will be maximum”.
Ques 8. A series RLC circuit resonance at 1 MHZ. At frequency 1.1 MHZ the circuit impedance will be
Will depend on the relative amplitude of RLC
For Series RLC circuit if the frequency is greater than the resonance frequency than the circuit behaves as an inductive circuit.
Ques 9. The equivalent resistance between terminal X and Y of the network shown is
As the given circuit is Wheatstone bridge balance circuit
Therefore according to balanced equation of Wheatstone bridge
Req = 20||40
= (20 x 40)/(20 + 40)
Ques 10. Application of Thevenin’s theorem in a circuit result in
An ideal voltage source
An ideal current source
A current source and an impedance in parallel
A voltage source and an impedance in series✔
The Thevenin equivalent resistance Rs is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage Vs and internal resistance RT and this is called Thevenin equivalent circuit.
SSC JE electrical paper 2014 Morning shift with explanation
Ques 1. A lamp having mean spherical candle power of 800 is suspended at a height of 10 m. Calculate the illumination just before the lamp
Illumination in foot-candles on plane normal to axis of light = Candle-power/ (distance)2
E = Icosθ/d2
= 800/102 = 8 lux
Ques 2. Hydrogen is used in large alternator mainly to
Reduce eddy current losses
Reduce distortion of waveform
Cool the machine✔
Strengthen the magnetic field
Why is hydrogen used for Alternator cooling?
Hydrogen is least expensive, less weight, high thermal conductivity, less density and less viscosity. Less weight, less density & less viscosity attributes to its flow rate. High thermal conductivity helps in better heat exchange. Least expensive helps in balance sheets, more power in less investments.
In order to reduce high temperature of alternator hydrogen gas is used as a coolant. The coolant, Hydrogen gas is allowed to flow in a closed cyclic path around the rotor. Heat exchange takes place and the temperature of hydrogen gas increases, for better cooling of the rotor in next cycle it has to be cooled. Cooling of hydrogen gas is done by passing it through heat exchangers generally constituted with water. Now Hydrogen gas after cooling is allowed to pass through driers ( mainly silica gel which absorbs moisture) and allowed to pass again through the rotor.
Ques 3. Two wires A and B have the same cross-section and are made of the same material. RA = 800Ω and RB = 100Ω. The number of times A is longer than B is
None of the options is correct check Explanation
Since they are of the same material they will have the same resistivity.
Therefore their length now
800/100 = la/lb
la = 8lb
Therefore A is 8 times longer than B
Ques 4. In the circuit shown in the figure, find the transient current i(t) when the switch is closed at t = 0, Assume zero initial condition.
50 t e-0.5t
50 t e-5t
100 t e-5t✔
100 t e-0.5t
Ques5. The Ebers-Moll model is applicable to:
Ebers Moll model is a simple and elegant way of representing the transistor as a circuit model.This model is based on assumption that base spreading resistance can be neglected. Ebers Moll model is one of the classical models of BJT for small signals. This model is based on interacting diode junctions and is applicable to any transistor operating in active mode
Ques6. A DC voltmeter has a sensitivity of 1000Ω /volt. When it measures half full scale in 100 V range, the current through the voltmeter will be
Sensitivity of voltmeter (s) = full scale deflection of voltmeter/current passing through voltmeter
For 1 volt resistance is 1000 ohm as given in the question.
For 100 volt resitance = 1000 x 100 ohm
And thus full current in the meter = v/r
And therefore half full scale 1/2= 0.5ma
Ques 7. A delta-star transformer has a phase to phase voltage transformation ratio of a : 1 [delta phase: star phase]. The line to line voltage ratio of star-delta is given by
Ques 8. Which of the following motors can be run on A.C as well as D.C supply?
The universal motor is so named because it is a type of electric motor that can operate on AC or DC power.
It is a commutated series-wound motor where the stator’s field coils are connected in series with the rotor windings through a commutator.
It is often referred to as an AC series motor. The universal motor is very similar to a DC series motor in construction but is modified slightly to allow the motor to operate properly on AC power.
This type of electric motor can operate well on AC because the current in both the field coils and the armature (and the resultant magnetic fields) will alternate (reverse polarity) synchronously with the supply.
Ques9. The power factor of the circuit shown in figure:
Power factor of series RL circuit
Due to inductive circuit power factor is lagging
Ques 10. The power factor of the A.C circuit is given by:
In electrical engineering, the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit ” R/Z”.
SSC JE electrical paper 2014 Evening shift with explanation
Ques1. A stove element draws 15A when connected to 230 V line. How long does it take to consume one unit of energy?
P = E/T
Where E is the energy and T is time
T = E/P
1 Unit of energy is given as 1 Kwhr = 1000 Watt-hour
T = 1000/230 x 15 [since P = VI]
T = 0.29 hour
Ques 2. The Req for the circuit shown in figure is
In the above figure 6Ω and 3Ω are parallel therefore
(6 x 3)(6 + 3) = 2Ω
Now 2Ω and 2Ω, as well as 1Ω and 5Ω, are in series therefore
2Ω + 2Ω = 4Ω
5Ω + 1Ω = 6Ω
Ques3. The SI unit of conductivity is
Conductivity (or specific conductance) is a measure of its ability to conduct electricity. The SI unit of conductivity is mho per meter.
Ques 4. Calculate the voltage drop across 14.5 Ω Resistance
In the given figure the resistances are in series, so the same current I will flow across the three resistors.
Requ = 14.5Ω + 25.5Ω + 60Ω = 100Ω
Current I = V/R
= 200/100 = 2A
Voltage drop across the 14.5Ω resistance
V = IR
= 2 x 14.5 = 29 V
Ques 5. For the network shown in the figure, the value of current in 8Ω resistors is
4. 8 A
The resistance between the point A and C will be
RAB || (RAC + RBC)
AB = 20 x(12 + 8)/(20 + 12 + 8) = 10Ω
I = V/R = 48/10 = 4.8 A
Since AC and BC are parallel with AB, therefore, the current will divide equally into both side
hence the current through the Branch ABC is = I/2
= 4.8/2 = 2.4 A
Ques 6. A piece of oil-soaked paper has been inserted between the plates of a parallel plate capacitor. Then the potential difference between the plates will
The paper sheet is the poor conductor of electricity so it does not allow the flow of electric current or electric charges between two parallel plates. However, the paper sheet allows electric field through it. Therefore, the paper sheet placed between the parallel plates acts as the barrier for the electric current. Hence the potential difference between the plates will decrease.
Ques 7. The current drawn by a tungsten filament lamp is measured by an ammeter. The ammeter reading under steady-state condition will be ________ the ammeter reading when the supply is switched on
When a filament lamp is cold(when supply is switched off), it behaves differently than when it’s at its normal operating temperature. The resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold. and in the process the lamp current changes. In the case of a cold incandescent lamp (a filament lamp), the inrush current peaks at approximately ten times the steady-state operating current, but it lasts a relatively short duration, no more than a few cycles.
Therefore the ammeter reading under the steady-state condition is less than then ammeter reading when the supply is switched on.
Ques 8. Tesla is same as
The Tesla (symbol T) is the SI derived unit used to measure magnetic fields. One tesla is equal to one weber per square meter.
Ques 9. The unit of volume of resistivity is
Resistivity is the resistance of a unit volume of a material. In the metric system, the unit of length is the meter, and the area is the square meter. Thus, resistivity is measured in units of Ohm – meters squared per meter (Ohm-m2/m), often abbreviated as Ohm-m.
Ques 10. Four resistance 2Ω, 4Ω, 5Ω, 20Ω are connected in parallel. Their combined resistance is
Since the resistance is connected in parallel therefore
1/ Req = 1/2Ω + 1/4Ω + 1/5Ω + 1/20Ω
=10 + 5 + 4 + 1/20
Note:- The total equivalent resistor connected in parallel is always less than the smallest given resistance. Here in the question, the smallest resistance is 2Ω and in the given option only option 1 i.e 1 Ω is less than the smallest resistance
so in the above case you can directly find the answer without any calculation.
SSC JE 2015 Electrical question paper Fully Solved
Ques 1. How much energy is stored in a 100 mH inductance when a current of 1A is flowing through it?
The energy stored in the magnetic field of an inductor can be expressed as
E = 1/2 L I2
E = energy stored (joules, J)
L = inductance (henrys, H)
I = current (amps, A)
1/2(100 x 10-3 x 1)
50 x 10-3 J
Ques 2. For the circuit shown below, find the resistance between points P & Q
Ques 3. The rate of change of current in a 4 H inductor is 2 Amps/sec. Find the voltage across inductor
8 V ✔
Here L = 4H and di/dt = 2 A
Hence voltage Across inductor
= V = L(di/dt)
= 4 x 2 = 8 V
Ques 4. Find the node voltage VA
6 V ✔
Ques 5. In a pure inductive circuit if the supply frequency is reduced to half the current will?
Be four times as high
Be reduced to half
Be reduced to one fourth
The amplitude of the current in a pure inductive circuit is given by
Im = Vm/ωl
And ω = 2πf
Im = Vm/2πfl
Im ∝ 1/f
Hence, when the frequency is halved, the amplitude of the current is doubled in a pure inductive circuit.
Ques 6. A 10 pole 25 Hz alternator is directly coupled to and is driven by 60 Hz synchronous motor then the number of poles in a synchronous motor is?
24 poles ✔
None of the above
Number of poles of alternator Pa = 10
F = 25 Hz (alternator)
F = 60 Hz (motor)
Then the number of poles of motor Pm =?
Since synchronous motor is directly coupled hence
Synchronous speed of an alternator = Synchronous speed of the motor
(120 x 25)/10 = (120 x 60)/ Pm
Pm = 24
Ques 7. When a source is delivering maximum power to the load the efficiency will be?
Below 50 %
Maximum Power Transfer theorem
The maximum power is transferred when the load resistance RL is equal to internal resistance Rth or the equivalent resistance Rth.
RL = Rth
Under this condition, the same amount of power is dissipated in the internal resistance Rth and hence the efficiency is 50%.
Ques 8. In the Maxwell Bridge as shown in the figure below the value of resistance Rx and inductance Lx of a coil is to be calculated after balancing the bridge. The component value is shown in the figure at balance the value of Rx and Lx will respectively be
37.5 ohms, 75 mH
75 ohm, 75 mH
375 ohm, 75 mH ✔
75 ohm, 150 mH
A Maxwell bridge is a modification to a Wheatstone bridge used to measure an unknown inductance (usually of low Q value) in terms of calibrated resistance and inductance or resistance and capacitance.
The balanced condition for Maxwell bridge is given as
Z1Z4 = Z2Z3
Here Z1 = Rx + jωLx
Z2 = 200Ω
Z3 = 750Ω
Z4 = R4 ||(1/ωC4)
Ques 9. The internal resistance of a voltage source is 10 ohm and has 10 volts and its terminals. Find the maximum power that can be transferred to the load
2.5 W ✔
For maximum power to be transferred from a source to a load resistance, the value of load resistance should be equal to the internal resistance of the source.
RL = R = 10Ω
current through the circuit = 10/(10+10) = 0.5 A
power transferred = I2R = 0.25 x 10 = 2.5 W
Ques 10. As the load is increased the speed of the DC shunt motor is
For a dc shunt motor. ﬁeld current is practically constant and thus the torque is directly proportional to the armature current. Hence the torque-armature current characteristic is a straight line passing through the origin. However, due to the armature reaction under the loaded conditions of the machine. the flux decreases slightly with the load.
Since flux is practically constant for DC shunt motor.The speed of DC shunt motor is
N = K (V – IaRa)
Armature current increases as the load on the motor is increased. Thus, the speed of the dc shunt motor will fall slightly with the armature current. As the speed characteristic is only slightly drooping, the dc shunt motor is normally regarded as a constant speed motor.