Exercises on Electric Generators and Electromotive Force –

LETTER C

Using the graph, we will write two functions using the equation of electrical generators:

\(U=\varepsilon-r\cdot i\)

{ 20=ε-5∙r 14=ε-8∙r

Next, we will solve the system by multiplying the second equation by -1 and adding the equations:

{ 20=ε-5∙r -14=-ε+8∙r

\(20-14=-5\cdot r+8\cdot r\)

\(6=3\cdot r\)

\(r=\frac{6}{3}\)

\(r=2\ \Omega\)

Now, we substitute the resistance value in the first equation and obtain the emf:

\(20=\varepsilon-5\cdot r\)

\(20=\varepsilon-5\cdot2\)

\(20=\varepsilon-10\)

\(\varepsilon=20+10\)

\(\varepsilon=30\ V\)

Finally, we will find the value of the short-circuit current that occurs when the electrical voltage is zero. To do this, we will again use the equation for electrical generators:

\(U=\varepsilon-r\cdot i\)

\(0=30-2\cdot i\)

\(-30=-2\cdot i\)

\(30=2\cdot i\)

\(i=\frac{30}{2}\)

\(i=15\ A\)