Find the following currents. The current I1 through the resistor of resistance R1 = 15.0 Ω . The current I2 through the resistor of resistance R2 = 45.0 Ω . The current I3 through the resistor of resistance R3 = 20.0 Ω . The current I4 through the resistor of resistance R4 = 25.0 Ω .

Explanation:

Assuming the circuit is the shown in the attached image, where:

We can use the 2nd Kirchoff Law (or Voltage Law) for each section of the circuit:

1st section:

2nd section:

Now, according to Ohm's Law, the voltage is equal to the multiplication of the current and the resistance :

Then:

1st section:

(1)

2nd section:

(2)

In this case the same current that gous trhough is the same that goes through , since they are in series.

And according to 1st Kirchoff Law (or Current Law):

(3)

At this point we have a system with three equations. Let's begin by substituting (3) in (1):

(4)

Applying common factor :

(5)

Isolating from (2):

(6)

Rewriting with the known values:

(7)

(8)

Substituting (8) in (5):

(9)

Isolating :

(10)

Rewritting with the known values;

(11)

(12)

Then:

Substituting (12) in (3):

(13)

(14)

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

number of electrons for neutral cl