RBJ1AO-M10 ΚΩRcVcc = + 15 VJ21 ΚΩ2N3904Note: Remove jumper wires J1 and J2 to insertthe ammeter and measure the base andcollector currents, respectively.Figure 6-3. BJT NPN switching circuit. b. If node "A" is connected to Vcc: saturation conditionShow your work on the spaces below: state the formula used, substitute given and write thefinal answer on the blank space provided. Hint: use KVL for the two loops; VLED = 1.7 V and VCE =OVInput side: Base-emitter loop |KVL Equation:Solve for /B (sat):

For the given BJT NPN switching it is asked to calculate the base and collector currents along with…

A Q- J1 RB 10 ΚΩ Rc Vcc = + 15 V J2 1 ΚΩ 2N3904 Note: Remove jumper wires J1 and J2 to insert the ammeter and measure the base and collector currents, respectively. Figure 6-3. BJT NPN switching circuit.

A Q- J1 RB 10 ΚΩ Rc Vcc = + 15 V J2 1 ΚΩ 2N3904 Note: Remove jumper wires J1 and J2 to insert the ammeter and measure the base and collector currents, respectively. Figure 6-3. BJT NPN switching circuit.

Electric Motor Control

10th Edition

ISBN:9781133702818

Author:Herman

Publisher:Herman

Chapter32: Two-speed, One-winding (consequent Pole) Motor Controllers

Section: Chapter Questions

Problem 6SQ: Describe the operation of Figure 325 (B) by adding jumper wire A to the original circuit.

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Concept explainers

Speed control of motor

Speed control by DC motor-speed controller is required to realize variable speed drive. Whenever a motor drives a particular load, it is called drive, and variable speed drive refers to a load requiring variable speed. All this is obtained by a DC motor speed controller.

Question

RB
J1
AO-M
10 ΚΩ
Rc
Vcc = + 15 V
J2
1 ΚΩ
2N3904
Note: Remove jumper wires J1 and J2 to insert
the ammeter and measure the base and
collector currents, respectively.
Figure 6-3. BJT NPN switching circuit.

b. If node "A" is connected to Vcc: saturation condition
Show your work on the spaces below: state the formula used, substitute given and write the
final answer on the blank space provided. Hint: use KVL for the two loops; VLED = 1.7 V and VCE =
OV
Input side: Base-emitter loop |
KVL Equation:
Solve for /B (sat):

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