what r the 2 main characteristics of a shunt connected motor?

http://www.answers.com/topic/direct-current-motor

sorry cant find what i'm looking for. anything else?
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COMPARE THE STARTING TORQUE OF A SHUNT AND SERIES MOTOR?
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COMPARE THE NO LOAD SPEEDSOF SHUNT AND SERIES CONNECTED MOTORS?

Hmmm. That specifically was in the article. I wonder how you missed it. Series motors are used for high torque applications, shunt motors are used for level load applications. I suggest reread the link I sent you.

sorry don't know how i missed itbut they only give one characteristic i still need another one

what is IT ETT ABOUT

A load draws 10 A of current from a 120-V source. The voltage leads the current by 36.9 degrees. Determine the following:
(a) _________ Resistive current
(b) __________ Reactive current

The main two characteristics of a shunt connected motor are:

1. Constant speed: Shunt connected motors maintain, almost, a constant speed regardless of the load. This characteristic makes them suitable for applications where constant speed is required, such as lathes, blower fans, and machine tools.

2. High starting torque: Shunt connected motors have relatively high starting torque, which makes them suitable for various industrial applications.

Now, for the comparison of starting torque and no-load speeds between shunt and series connected motors:

Starting Torque:
- Shunt motor: High starting torque
- Series motor: Extremely high starting torque

No-load Speeds:
- Shunt motor: Almost constant speed, regardless of the load
- Series motor: No-load speed is very high and can be dangerously unstable. Hence, they are never operated without a load.

As for the last question:

A load draws 10 A of current from a 120-V source. The voltage leads the current by 36.9 degrees.

We can use the cosine of the angle to find the resistive and reactive currents:

Resistive current (I_R) = I * cos(angle)
Reactive current (I_X) = I * sin(angle)

where I = total current, and the angle = 36.9 degrees

(a) Resistive current:
I_R = 10A * cos(36.9) = 10A * 0.8 = 8 A

(b) Reactive current:
I_X = 10A * sin(36.9) = 10A * 0.6 = 6 A

To find the resistive and reactive currents, we can use the concept of power triangle and trigonometric calculations. Here's how you can calculate them step by step:

Step 1: Convert the given voltage and current values to complex notation:
The given voltage, V = 120 V
The given current, I = 10 A

Representing these values in complex notation:
V = 120∠0° V
I = 10∠36.9° A

Step 2: Calculate the apparent power (S):
Apparent power (S) is given by the product of voltage and current:
S = V * I

S = (120∠0°) * (10∠36.9°) A
S = 1200∠36.9° VA

Step 3: Calculate the real power (P):
Real power (P) is given by the product of apparent power and the cosine of the phase angle (36.9°):
P = S * cos(36.9°)

P = 1200∠36.9° * cos(36.9°)
P = 960 W (rounded to the nearest whole number)

Step 4: Calculate the reactive power (Q):
Reactive power (Q) is given by the product of apparent power and the sine of the phase angle (36.9°):
Q = S * sin(36.9°)

Q = 1200∠36.9° * sin(36.9°)
Q = 720 VAR (rounded to the nearest whole number)

Step 5: Calculate the resistive current (I_r):
The resistive current (I_r) is given by the real power divided by the voltage:
I_r = P / V

I_r = 960 W / 120 V
I_r = 8 A

Step 6: Calculate the reactive current (I_x):
The reactive current (I_x) is given by the reactive power divided by the voltage:
I_x = Q / V

I_x = 720 VAR / 120 V
I_x = 6 A

Summary:
(a) The resistive current is 8 A.
(b) The reactive current is 6 A.

To determine the resistive and reactive currents in a circuit, you can use the concept of phasors and complex numbers.

First, let's start by calculating the apparent power (S) in the circuit using the given values of current (I) and voltage (V).

S = V * I

Given that the current is 10 A and the voltage is 120 V:

S = 120 V * 10 A = 1200 VA

The apparent power (S) represents the vector sum of the real power (P) and the reactive power (Q). Since we know that the power factor (PF) is the cosine of the phase angle between the voltage and current, we can calculate the real power as:

P = S * PF

The power factor (PF) can be calculated as:

PF = cos(angle)

Given that the phase angle is 36.9 degrees, the power factor (PF) can be calculated as:

PF = cos(36.9 degrees)

Using a scientific calculator or trigonometric tables, you can find that cos(36.9 degrees) is approximately 0.7986.

Therefore, the real power (P) is:

P = 1200 VA * 0.7986 ≈ 957.12 W

Now, to determine the reactive power (Q), we can use the following formula:

Q = sqrt(S^2 - P^2)

Plugging in the values, we get:

Q = sqrt((1200 VA)^2 - (957.12 W)^2)

Using a calculator, we find that Q ≈ 771.82 VAR

To calculate the resistive current and reactive current, we need to know the impedance (Z) of the load. If the load is purely resistive, then the reactive current will be zero. However, if the load has a reactive component, then we can calculate the resistive current and reactive current using the following formulas:

Resistive current (I_R) = P / V

Reactive current (I_X) = Q / V

Since the given information does not provide the impedance or specify the nature of the load (whether it's purely resistive or has a reactive component), we cannot determine the exact values of the resistive and reactive currents.