MM0704, Lesson 4
If five vacuum tubes, each drawing 0.12 amps of current, are connected to the secondary, the total secondary current
will be 5 x 0.12 or 0.6 amps. The primary current will be
The energy delivered to and supplied by the primary is: Ip Ep = 0.027 x 110 = 2.97 W or approximately 3 W. As a
check, note that each tube takes 0.12 amps at 5 V; each tube is using: 5 x 0.12 = 0.6 W. Five of these tubes require: 5 x
0.6 = 3 W.
Since this transformer, which is rated at 10 V-amp (which is 10 W for a resistive load), is delivering only 3 W, it is
loaded at: 3 + 10 x 100 percent = 30 percent of its rated load.
Impedance Ratio. Impedance (Z)
to E divided
by I. Therefore,
the ratio is then
Substituting from equations in the second paragraph in this section.