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For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknowns (current Ib2 neglected), Let me try: Vb1-Vx=I5*R5 I5=IC4+Vb1/R3 IC4=Ue...
Answer
#3: Post edited
by
LvW
·
2024-05-12T14:23:24Z (7 months ago)
- For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknowns (current Ib2 neglected),
- Let me try:
- * Vb1-Vx=I5*R5
- * I5=IC4+Vb1/R3
- * IC4=Ue4/R4
- * Ue4=VB2-0.7
- These 4 equations (simple insertion procedure) result in:
- * Vb1-Vx=R5[(Vb2-0.7)/R4 + Vb1/R3]
- * Now we can Vb2 express as a function Vx (voltage division) and we have an equation with the **two unknowns Vb1 and Vx**.
- * A **second equation for Vb1 and Vx** can be derived from
- (V1-Vx)/R10=I10=I5+(Vx/Rx) with I5=(Vb1-Vx)/R5.
* The resistor Rx is the total resistance between Vx and ground.
- For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknowns (current Ib2 neglected),
- Let me try:
- * Vb1-Vx=I5*R5
- * I5=IC4+Vb1/R3
- * IC4=Ue4/R4
- * Ue4=VB2-0.7
- These 4 equations (simple insertion procedure) result in:
- * Vb1-Vx=R5[(Vb2-0.7)/R4 + Vb1/R3]
- * Now we can Vb2 express as a function Vx (voltage division) and we have an equation with the **two unknowns Vb1 and Vx**.
- * A **second equation for Vb1 and Vx** can be derived from
- (V1-Vx)/R10=I10=I5+(Vx/Rx) with I5=(Vb1-Vx)/R5.
- * The resistor Rx is the total resistance between Vx and ground.
- * Last step: Solve the system with two equations for two unknowns.
#2: Post edited
by
LvW
·
2024-05-12T14:21:48Z (7 months ago)
For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknown (Current Ib2 neglected),- Let me try:
- * Vb1-Vx=I5*R5
- * I5=IC4+Vb1/R3
- * IC4=Ue4/R4
- * Ue4=VB2-0.7
- These 4 equations (simple insertion procedure) result in:
- * Vb1-Vx=R5[(Vb2-0.7)/R4 + Vb1/R3]
- * Now we can Vb2 express as a function Vx (voltage division) and we have an equation with the **two unknowns Vb1 and Vx**.
- * A **second equation for Vb1 and Vx** can be derived from
- (V1-Vx)/R10=I10=I5+(Vx/Rx) with I5=(Vb1-Vx)/R5.
- * The resistor Rx is the total resistance between Vx and ground.
- For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknowns (current Ib2 neglected),
- Let me try:
- * Vb1-Vx=I5*R5
- * I5=IC4+Vb1/R3
- * IC4=Ue4/R4
- * Ue4=VB2-0.7
- These 4 equations (simple insertion procedure) result in:
- * Vb1-Vx=R5[(Vb2-0.7)/R4 + Vb1/R3]
- * Now we can Vb2 express as a function Vx (voltage division) and we have an equation with the **two unknowns Vb1 and Vx**.
- * A **second equation for Vb1 and Vx** can be derived from
- (V1-Vx)/R10=I10=I5+(Vx/Rx) with I5=(Vb1-Vx)/R5.
- * The resistor Rx is the total resistance between Vx and ground.
#1: Initial revision
by
LvW
·
2024-05-12T14:21:09Z (7 months ago)
For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknown (Current Ib2 neglected), Let me try: * Vb1-Vx=I5*R5 * I5=IC4+Vb1/R3 * IC4=Ue4/R4 * Ue4=VB2-0.7 These 4 equations (simple insertion procedure) result in: * Vb1-Vx=R5[(Vb2-0.7)/R4 + Vb1/R3] * Now we can Vb2 express as a function Vx (voltage division) and we have an equation with the **two unknowns Vb1 and Vx**. * A **second equation for Vb1 and Vx** can be derived from (V1-Vx)/R10=I10=I5+(Vx/Rx) with I5=(Vb1-Vx)/R5. * The resistor Rx is the total resistance between Vx and ground.