AIIMS CRE Electrical JE/AE Mock Test- 3 - NEET MCQ

Hence, the correct option is (b)

Any device through which we give any data to a computer or personal computer, it is called input and the device is called input device. Some examples of this are: keyboard, mouse, scanner, all these input devices we input something inside the computer.ng inside the computer with input devices.

Semaphore is an integer variable that is used to prevent threads or processes from accessing a single resource. Semaphore is simply a variable that is non-negative and shared between threads. A semaphore is a signaling mechanism, and a thread that is waiting on a semaphore can be signaled by another thread. It uses two atomic operations, wait, and signal for the process synchronization.

Power consumed

230 X 15 X t = 1000 [1 unit = 1000 W - h ]

t = 0.29 hour

If Quantity is voltage

So, by observing Peak to peak voltage of sine wave, we can get the RMS value of that quantity.

From the short circuit test of the Transformer we will find out full load copper loss and equivalent circuit parameter & voltage regulation.
From full load copper loss we can find out copper losses at any fraction of load by using the following formula.
W_cu at x% full load = x² [W_cu full load]

We have, intrinsic carrier concentration, 
ηi∞e^-Eg/2kT
And, hole concentration (by mass action law) is, p=ni²/n⇒∞e^-Eg2kT
Thus, E_gA>E_gB
⇒pA<pB
Hence, hole concentration in semiconductor A is less than that of B.

When 4Ω is in the circuit,

A growler is used to check the armature for shorts, opens and grounds.

E= Eo – Ep

Reverse saturation current (IS) of diode increases with increase in the temperature, the rise is 7%/ºC for both germanium and silicon and approximately doubles for every 10ºC rise in temperature. Thus if we kept the voltage constant, as we increase temperature the current increases.

Since I_sh = 220/ 50 = 4.4 A
I_a = 80-4.4 = 75.6 Amp
E_b= V - I_aR_a
= 220- (75.6 x 0.1) = 212.44
Driving Power E_b.I_a = 212.44x75.6
= 16060 N-m
Output Power = Driving Power -  Iron and Friction Losses
= 16060-1600
= 14460 W
Armature Torque
T_a = E_bI_a/2πN   if N in r.p.s
T_a = (60E_bI_a)/(2πN)  if N in r.p.m
so; Ta = (9.55E_bI_a)/N   if N in r.p.s
= (9.55 x 212.44 x 75.6)/800
= 191.7 N-m
Since the whole of armature torque is not available for doing useful work, a certain part is required to overcome iron & friction losses
Hence Lost torque = 9.55x (1600/800)
=  19.1 N-m
Hence Useful Shaft torque = T_a - T_lost
= 191.7 - 19.1
= 172.6 N-m

Both hysteresis and eddy current losses depend upon A. maximum flux density B_m in the core and B. supply frequency f. Since transformers are connected to constant voltage and constant-frequency supply, both B_m and f  are constant. Hence iron or core losses in a transformer are practically the same at all loads.

Generally, superposition theorem is used to find voltages and currents. And it is applicable only for linear, bilateral elements.
Superposition theorem is not applicable for power calculation.
Note:

• The superposition theorem states that "in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently."
• The superposition theorem applies only when all the components of the circuit are linear, which is the case for resistors, capacitors, and inductors it is not applicable to networks containing nonlinear elements.

Multiplying power