Example of problem solution

Example 3 The electric circuit consists of two sources of EMF e₁=20 V, e₂=5 V and three resistors R _1, R ₂=19W and R ₃=10W. A current I ₁=0, 2 А flows through the inner resistances of the sources r ₁=2W, r ₂=1W and through the resistor R ₁, at the direction, shown in the picture.

Find: 1) resistance R ₁ and a current, which flows through the resistors R ₂ and R ₃;

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

Input data: e 1=20 V e 2=5 V r 1=2 W r 2=1 W R 2=19 W R 3=10 W I 1=0, 2 А
Find: I 2, I 3, R 1, 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 I ₁ is set, and let’s choose directions of currents I ₂ and I ₃, 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 I ₃ from the equation (4) and insert to the equation (6)

.

Whence

Minus-sign in the sense of a current I ₂ means, that the current’s direction I ₂ has been chosen reverse to the acting. In reality current I ₂ runs from junction В to the junction А.

From the equation (4) we search out I ₃:

From the equation (5) we search out R ₁

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; I ₃ = 0, 3 А; R ₁=83 W, .

Individual tasks for PROBLEM 1.5.