Example of problem solution

Example 3 The electric circuit consists of two sources of EMF e₁=20V, e₂=5V and three resistors R_1, R₂=19W and R₃=10W. A current I₁=0,2А flows through the inner resistances of the sources r₁=2W, r₂=1W and through the resistor R₁, at the direction, shown in the picture.

Find: 1) resistance R₁ and a current, which flows through the resistors R₂ and R₃;

2) the potential difference between the points А and В.

Input data: e1=20 V e2=5 V r1=2 W r2=1 W R2=19 W R3=10 W I1=0,2 А
Find: I2, I3, R1, DjВA–?

Solution:

1. Let’s use the Kirchhoff’s rules for the solution of the branched chain.

In order to find one magnitude of resistance and two magnitudes of a current, it’s necessary to make three equations. Before compiling the equations it’s necessary arbitrarily to choose:

a) direction of currents (if they aren’t set in the condition); b) direction of path-tracing.

The direction of a current I₁ is set, and let’s choose directions of currents I₂ and I₃, as it’s shown on the scheme. Let’s agree to trace-paths clockwise (dashed line on the scheme). The given scheme has two junctions: А and В. In order to compile the equations with the first Kirchhoff’s rule it’s necessary to take into account, that a current, which flows in the junction, enters to the equation with plus-sign, and it is necessary to write a current, which flows out of the junction with minus-sign.

With the first Kirchhoff’s rule for the junction А

(1)

It is no sense to compile the equation for the junction В, as it reduces to the equation (1).

We’ll get two more necessary equations with the second Kirchhoff’s rule. It is necessary to follow the sign rules: a) the voltage drop (product IR or Ir) enters to the equation with plus-sign, if the current’s direction coincides with the direction of path-tracing, in other case – with minus-sign; b) EMF enters to the equation with plus-sign, if it enlarges the potential in the direction of path-tracing (pass through the source from minus to plus), in other case – with minus-sign.

With the second Kirchhoff’s rule for the closed loop and :

; (2)

. (3)

The set magnitudes inserted in the equation (1), (2), (3), we’ll get the set of equations

Let’s find I₃ from the equation (4) and insert to the equation (6)

.

Whence

Minus-sign in the sense of a current I₂ means, that the current’s direction I₂ has been chosen reverse to the acting. In reality current I₂ runs from junction В to the junction А.

From the equation (4) we search out I₃:

From the equation (5) we search out R₁

W.

2. The difference of potential U=Dj_A,B=j_B–j_A can be found, if we use Ohm’s law for Non-uniform site of a chain (subcircuit) in a proper way, for a example

(7)

In Ohm’s law it’s taken into consideration, that positive direction of current’s strength coincides with the direction of foreign forces work of the source, which fits the enlarging of the potential. Then the required potential difference is

We make the calculations

.

Results: I₂ = – 0,1 A; I₃ = 0,3 А; R₁=83 W, .

Individual tasks for PROBLEM 1.5.

BRANCHED CIRCUITS

To compose the scheme from three adjacent branches, which are shown in the picture 4. The numbers of branches, EMF of sources e_i, the inner resistance of sources r_і, the external resistance of branches R_i (or the current І_і, which flows along one of the branches from the point А to В) set with the variants in the table. 1.5. Find: 1) magnitudes of quantities, which are indicated in the last column of the Table 1.5;

2) the potential difference from the points А and В.

Example of the scheme, which corresponds 25^th variant is shown in the picture 4 а.

TABLE OF TASK VARIANTS

Table 1.5

Variant	The branch number	εi , V	ri , W	Ri , W	Іі , А	Find
	1, 2, 3	ε1 =11, ε2 =4, ε3 =6	r1= r2 = r3 =0	R1 =25, R2 =50, R3 =10	–	І1, І2, І3
	4, 5, 6	ε4 =9, ε5 =10	r4 =1, r5 =2	R4 =19, R5 =38	І6 =0,1	І4, І5, R6
	1, 2, 4	ε1 =16, ε2 =5, ε4 =7	r1= r2 = r4 =0	R2 =30, R4 =50	І1 =0,4	І2, І4, R1
	5, 4, 1	ε1 =9, ε4 =6, ε5 =2	r1= r4 = r5 =0	R4 =50, R5 =10	І1 =0,2	І4, І5, R1
	1, 2, 6	ε1 =10, ε2 =8	r1 =2, r2 =1	R1 =8, R2 =19, R6 =60	–	І1, І2, І6
	3, 2, 1	ε2 =4, ε3 =5	r1= r2 = r5 =0	R1 =30, R2 =40, R3 =20	І1 =0,1	І2, І3, ε1
	1, 4, 6	ε1 =8, ε4 =2	r1 =2, r4 =1	R1 =18, R4 =39, R6 =80	–	І1, І4, І6
	1, 4, 2	ε2 =11, ε4 =7	r1= r2 = r4 =0	R1 =50, R2 =20, R4 =30	І1 =0,1	І2, І4, ε1
	2, 1, 3	ε1 =9, ε2 =8, ε3 =1	r1= r2 = r5 =0	R1 =50, R2 =20, R3 =10	–	І1, І2, І3
	4, 1, 5	ε4 =4, ε5 =2	r1= r4 = r5=0	R1 =25, R4 =50, R5 =10	І1 =0,4	І4, І5, ε1
	1, 3, 2	ε2 =16, ε3 =3	r1= r2 = r5 =0	R1 =70, R2 =20, R3 =10	І1 =0,1	І2, І3, ε1
	6, 4, 1	ε1 =3, ε4 =7	r1 =2, r4 =1	R1 =78, R4 =39	І6 =0,1	І1, І4, R6
	5, 4, 1	ε4 =4, ε5 =14	r1= r4 = r5=0	R1 =90, R4 =20, R5 =40	І1 =0,1	І4, І5, ε1
	4, 6, 5	ε4 =10, ε5 =5	r4 =2, r5 =1	R4 =33, R5 =19	І6 =0,3	І4, І5, R6
	1, 6, 4	ε1 =4, ε4 =3	r1 =2, r4 =1	R1 =18, R4 =9, R6 =60	–	І1, І4, І6
	4, 1, 6	ε1 =2, ε4 =12	r1 =3, r4 =2	R1 =97, R4 =18	І6 =0,1	І2, І4, R6
	4, 1, 5	ε1 =22, ε4 =8, ε5 =4	r1= r4 = r5=0	R1 =25, R4 =50, R5 =10	–	І1, І4, І5
	2, 1, 6	ε1 =20, ε2 =6	r2 =1	R1 =82, R2 =29, R6 =10	І1 =0,2	І2, І6, r1
	2, 3, 1	ε1 =19, ε2 =4, ε3 =5	r1= r2 = r3 =0	R2 =20, R3 =10	І1 =0,2	І2, І3, R1
	4, 1, 6	ε1 =13, ε4 =1	r4 =1	R1 =27, R4 =24, R6 =40	І1 =0,3	І4, І6, r1
	2, 1, 4	ε1 =12, ε2 =9, ε4 =5	r1= r2 = r4 =0	R1 =30, R2 =60, R4 =20	–	І1, І2, І4
	2, 1, 6	ε1 =8, ε2 =6	r1 =3	R1 =27, R2 =9, R6 =25	І2 =0,1	І1, І6, r2
	5, 1, 4	ε1 =19, ε4 =6, ε5 =2	r1= r4 = r5=0	R4 =50, R5 =10	І1 =0,2	І4, І5, R1
	1, 6, 2	ε1 =18, ε2 =15	r1 =2, r2 =1	R1 =58, R2 =9, R6 =30	–	І1, І2, І6
	4, 1, 2	ε2 =4, ε4 =2	r1= r2 = r4 =0	R1 =50, R2 =20, R4 =80	І1 =0,2	І2, І4, ε1
	1, 6, 5	ε1 =8, ε5 =6	r1 =2, r5 =3	R1 =8, R5 =12, R6 =10	-	І1, І5, І6
	2, 4, 5	ε2 =8	r2 =2, r4 =1, r5 =5	R2 =18, R4 =14, R5 =25	І4=0,2, І5=0,3	І2, ε4, ε5
	3, 6, 4	ε3 =36, ε4 =9	r3 =2, r4 =1	R3 =16, R4 =8	І6 =0,5	І4, І3, R6
	3, 1, 5	ε3 =40, ε5 =30	r1=r5=2, r3=5	R3 =35, R1 =28, R5 =28	І1 =0,7	І5, І3, ε1
	2, 3, 4	ε2=20, ε4=40, ε3=10	r2=10, r4=15, r3 =5	R2 =110, R4 =105	І3 =0,2	І4, І2, R3