Transcript Diapositiva 1

Calcular las corrientes de malla y los voltajes y corrientes que se piden para cada circuito.
3K
4K
Circuito 1
4V
A
C1.1. Corriente en la fuente de 10 V.
Método de mallas. Ejercicios.
C1.2. Corriente en la resistencia de 3K.
C1.3. Corriente en cada una de las resistencias de 1K en paralelo.
1K
10 V
5K
2K
2K
B
A
4V
C1.4. Caída de tensión desde A hasta B.
Circuito 2
C2.1. Corriente en la resistencia de 3K.
8K
5K
2V
C2.2. Corriente en la resistencia de 2K.
2K
3K
C2.3. Corriente en la resistencia de 8K.
2V
C2.4. Caída de tensión desde A hasta B.
B
A
20 V
10 V
3.5 K
Circuito 3
C3.1. Corriente en la resistencia de 1K compartida.
1K
1K
0.5 K
C3.2. Corriente en la fuente de 20 V.
C3.3. Corriente en la resistencia de 1K de la derecha.
10 V
C3.4. Caída de tensión desde A hasta B.
B
1
Calcular las corrientes de malla y los voltajes y corrientes que se piden para cada circuito.
A
A
Circuito 4
10 V
10 V
Método de mallas. Ejercicios.
C4.1. Corriente en la resistencia de 7K.
7K
2K
3K
1K
C4.2. Corriente en la fuente de 5 V.
C4.3. Corriente en la resistencia de 2K.
5V
B
C4.4. Caída de tensión desde A hasta B.
4K
1K
A
Circuito 5
C5.1. Corriente en la resistencia de 9K.
4K
C5.2. Corriente en la resistencia de 1K.
5V
C5.3. Corriente en la fuente de 5 V.
9K
C5.4. Caída de tensión desde A hasta B.
3K
2K
B
2
SOLUTIONS
4K
3K
Rp  1 K
4V
A
Circuit 1
MESH ANALYSIS. EXERCISES
C1.1. Current in the 10 V source?
1K
i1
10 V
1K
C1.2. Current in the 3K resistance?
i2
C1.3. Current in each of the 1K resistances
connected in paralell?
5K
2K
2K
C1.4. Potential drop from point A to point B?
Mesh currents
B
Equation
of the
circuit
 5  1  i1  10

     
  1 10   i2   4 
C1.1. Current in the 10 V source?
C1.2. Current in the 3K resistance?

5 1
 1 10
1 
10  1
4 10
2 
5 10
1 4
i1 
1
 2.12 mA

i2 
2
 0.61 mA

i10V  i1  2.12 mA
i2
i3K  i2  0.61mA
C1.3. Current in each of the 1K resistances
connected in paralell?
i1K 
i2
 0.305 mA
2
i2 / 2
i2
1K
(current divider)
2K
2K
C1.4. Potential drop from point A to point B?
VAB
i
 3i2  4  2 2  1.55 V
2
VAB  1i1  i2   5i2  1.55 V
B
VA  VB 
3
i2 / 2
SOLUTIONS
Circuit 2
A
MESH ANALYSIS. EXERCISES
4V
C2.1. Current in the 3K resistance?
8K
5K
2V
C2.2. Current in the 2K resistance?
2K
3K
i1
i2
C2.3. Current in the 8K resistance?
2V
C2.4. Potential drop from point A to point B?
B
Equation
of the
circuit
 10  2   i1   4  2   2 

    
   
  2 10   i2   2  2   4 
C2.1. Current in the 3K resistance?
C2.2. Current in the 2K resistance?
C2.3. Current in the 8K resistance?
Mesh currents

10  2
 2 10
1 
2 2
4 10
2 
i1 
1
 0.29 mA

i2 
2
 0.46 mA

10 2
2 4
i3K  i1  0.29 mA
i2 K  i1  i2  0.29  0.46  0.17 mA
(Same direction as i2)
i8K  i2  0.46 mA
C2.4. Potential drop from point A to point B?
VAB  5i1  2  2i1  i2   3.12 V
VAB  4  3i1  3.12 V
VA  VB 
4
SOLUTIONS
A
MESH ANALYSIS. EXERCISES
20 V
10 V
3.5 K
C3.1. Current in the 1K resistance (that on the middle) ?
1K
0.5 K
i1
Circuit 3
1K
i2
C3.2. Current in the 20 V source?
C3.3. Current in the 1K resistance (that on the right) ?
C3.4. Potential drop from point A to point B?
10 V
B
Mesh currents
Equation
of the
circuit
 5  1  i1    20    20

    
  

  1 2   i2  10  10  20 

5 1
1 2
1 
 20  1
20 2
2 
C3.1. Current in the 1K resistance (that on the middle)? i1K  i1  i2  11.1 mA
5  20
 1 20
i1 
1
 2.22 mA

i2 
2
 8.89 mA

The same direction as i2,
opposite direction as i1.
C3.2. Current in the 20 V source? i20V  i1  2.22 mA This minus sign means that the direction is
C3.3. Current in the 1K resistance (that on the right)?
i1K  i2  8.89 mA
C3.4. Potential drop from point A to point B?
VAB  10  1i1  i2   1.11V
VAB  1i2 10  1.11V
VA  VB 
5
SOLUTIONS
A
Circuit 4
A
10 V
MESH ANALYSIS. EXERCISES
10 V
C4.1. Current in the 7K resistance?
C4.2. Current in the 5 V source?
2K
7K
i1
3K
i2
1K
C4.3. Current in the 2K resistance?
C4.4. Potential drop from point A to point B?
5V
B
Mesh
currents
i3
4K
Equation
of the
circuit
 10  7  1  i1   10   10 

  
  
  7 10 0   i2   10  5    15 
 1 0
5   i3    5    5 

10
  7
1
C4.1. Current in the 7K resistance? i7 K  i1  i2  0.306mA
C4.2. Current in the 5 V source?
C4.3. Current in the 2K resistance?
i5V  i2  i3  4.490mA
i1 
i2 
 7 1
10
0
0
5
1
 3.980 mA

2
 4.286 mA

10
i3 
 7 1
1  15 10
5 0
0
5
The same direction as i2,
opposite direction as i1.
The same direction as i2,
opposite direction as i3.
3
 0.204 mA

10
10
1
2   7
15
0
1  5
5
10
 7 10
3   7
10
15
1
0
5
i2 K  i1  3.980 mA
C4.4. Potential drop from point A to point B?
VAB  10  3i2  2.86 V
VAB  10  2i1  4i3  2.86 V
VA  VB 
6
SOLUTIONS
Circuit 5
1K
A
MESH ANALYSIS. EXERCISES
C5.1. Current in the 9K resistance?
4K
9K
C5.2. Current in the 1K resistance?
i2
C5.3. Current in the 5 V source?
5V
C5.4. Potential drop from point A to point B?
i1
i3
3K
Equation
of the
circuit
2K
 16  4  3   i1   0 

   
0   i2    5 
 4 5
3 0
5   i3    5 

Mesh
currents
B
16
  4
3
i2 
4 3
5
0
0
5
1
 0.091 mA

i1 
2
 1.073 mA

0
i3 
4 3
1   5
5
0
5
0
5
3
 0.946 mA

16
0
3
2   4
5
0
3 5
5
4
0
5
0
5
5
16
C5.1. Current in the 9K resistance?
i9 K  i1  0.091mA
C5.2. Current in the 1K resistance?
i1K  i2  1.073mA
C5.3. Current in the 5 V source?
i5V  i2  i3  2.019mA
C5.4. Potential drop from point A to point B?
3  4
3
The same direction as i2,
opposite direction as i3.
VAB  1i2  1.07 V
VAB  9i1  2i3  1.07 V
VA  VB 
7
For you to solve
Circuit 6
A
3K
MESH ANALYSIS. EXERCISES
4K
2K
C6.1. Potential drop from point A to point B?
C6.2. Potential drop from point B to point C?
C6.3. Potential drop from point A to point C?
C6.4. Equivalent resistance of the set?
4K
3K
(Note that this set is neither a series nor
a parallel association).
B
C
4.5 V
8