Practice Problems (KEY)

1. Answer the following questions about the devices shown in the table below.

Device	Current	Voltage
A	3 \(\mathrm{A}_{RMS}\)	190 \(\mathrm{V}_{RMS}\)
B	8 \(\mathrm{A}_{RMS}\)	190 \(\mathrm{V}_{RMS}\)
C	3 \(\mathrm{A}_{RMS}\)	190 \(\mathrm{V}_{RMS}\)
X	10 \(\mathrm{A}_{RMS}\)	40 \(\mathrm{V}_{RMS}\)
Y	4 \(\mathrm{A}_{RMS}\)	40 \(\mathrm{V}_{RMS}\)
Z	5 \(\mathrm{A}_{RMS}\)	40 \(\mathrm{V}_{RMS}\)

a. Draw the block diagram to represent this set of devices.
b. Label circuit breaker locations and values.
c. Calculate total power supplied.

\[ P_{\mathrm{gen}} = I_S V_S = 18 \times 190 = 3.42\ \mathrm{kW} \]

2. Calculate the following for the power distribution system below:

a. Turns ratio of the transformer
b. Circuit breaker values for circuit breakers A, B, and C
c. Power the source produces

\[ a = \frac{V_1}{V_2} = \frac{35}{205} = 0.171 \]

\[ P_{\mathrm{flaps,out}} = 15\ \mathrm{hp} \cdot \frac{1492\ \mathrm{W}}{2\ \mathrm{hp}} = 11.19\ \mathrm{kW} \]

\[ P_{\mathrm{flaps,in}} = \frac{11{,}190\ \mathrm{W}}{0.62} = 18\ \mathrm{kW} \]

\[ I_S = \frac{18{,}000\ \mathrm{W}}{205\ \mathrm{V}} = 88\ \mathrm{A} \]

\[ I_C = I_{\mathrm{Flaps}} + I_{\mathrm{Avionics}} + I_{\mathrm{Nav}} = 88 + 12 + 12 = 112\ \mathrm{A} \]

\[ \mathrm{CB}\ I_C = 112 \times 1.1 = 123\ \mathrm{A} \rightarrow 125\ \mathrm{A} \]

\[ P_C = V_C I_C = (205\ \mathrm{V}_{RMS})(112\ \mathrm{A}) = 22.97\ \mathrm{kW} \]

\[ P_B = \frac{22{,}970\ \mathrm{W}}{0.8} = 28.7\ \mathrm{kW} \]

\[ I_B = \frac{28{,}700\ \mathrm{W}}{32\ \mathrm{V}} = 897\ \mathrm{A} \]

\[ \mathrm{CB}\ I_B = 897 \times 1.1 = 987\ \mathrm{A} \rightarrow 1\ \mathrm{kA} \]

\[ I_A = I_{\mathrm{Comms}} + I_{\mathrm{Cam}} + I_B = 5 + 2.3 + 897 = 904\ \mathrm{A} \]

\[ \mathrm{CB}\ I_A = 904 \times 1.1 = 994\ \mathrm{A} \rightarrow 1\ \mathrm{kA} \]

\[ P_S = V_S I_S = 32 \times 904 = 28.9\ \mathrm{kW} \]

3. T/F Buses work by delivering power to devices connected in series.

4. T/F A circuit breaker works the same as a fuse in that it opens when current exceeds the breaker’s rating.

5. The electrical system for a small reconnaissance aircraft is shown below. The aircraft has two buses, one at 160 \(\mathrm{V}_{RMS}\) and one at 50 \(\mathrm{V}_{RMS}\), powered by a 160 \(\mathrm{V}_{RMS}\) alternator.

a. What transformer turns ratio is required to drive the 50 \(\mathrm{V}_{RMS}\) bus?

\[ a = \frac{160}{50} = 3.2 \]

b. Given the following table of required currents, what circuit breaker ratings would you choose for breakers A, B, C, and D?

Equipment	Current (\(\mathrm{A}_{RMS}\))
Flaps	2.5
Avionics	4
Navigation	6
Comm	3.5
Camera	4.5

\[ I_D = 3.5 \Rightarrow \mathrm{CB}\ I_D = 3.85 \rightarrow 4\ \mathrm{A} \]

\[ I_C = 3.5 + 4.5 = 8 \Rightarrow \mathrm{CB}\ I_C = 8.8 \rightarrow 9\ \mathrm{A} \]

\[ I_B = \frac{8}{3.2} = 2.5 \Rightarrow \mathrm{CB}\ I_B = 2.75 \rightarrow 3\ \mathrm{A} \]

\[ I_A = 2.5 + 4 + 6 + 2.5 = 15 \Rightarrow \mathrm{CB}\ I_A = 16.5\ \mathrm{A} \]

6. The electrical system for a fighter is shown below. The aircraft has two buses, one at 80 \(\mathrm{V}_{RMS}\) and one at 20 \(\mathrm{V}_{RMS}\), powered by a 115 \(\mathrm{V}_{RMS}\) alternator.

a. What transformer turns ratios are required?

\[ a_{\mathrm{XFMR1}} = \frac{115}{80} = 1.4375 \quad a_{\mathrm{XFMR2}} = \frac{115}{20} = 5.75 \]

b. Given the following table, how much power must the alternator provide?

Equipment	Current (\(\mathrm{A}_{RMS}\))
Flaps	5
Landing Gear	8
Radar	4
Computer	2
IRS	5

\[ I_{20V} = 2 + 5 = 7 \Rightarrow I_{1,\mathrm{XFMR2}} = \frac{7}{5.75} = 1.21\ \mathrm{A} \]

\[ I_{80V} = 5 + 8 + 4 = 17 \Rightarrow I_{1,\mathrm{XFMR1}} = \frac{17}{1.4375} = 11.8\ \mathrm{A} \]

\[ I_S = 1.21 + 11.8 = 13\ \mathrm{A} \]

\[ P_{\mathrm{ALT}} = (13)(115) = 1.495\ \mathrm{kW} \]

Alternatively,

\[ P_{\mathrm{ALT}} = 80(5 + 8 + 4) + 20(2 + 5) = 1.5\ \mathrm{kW} \]

7. Repeat Problem 6 with the alternate configuration shown below.

\[ a_{\mathrm{XFMR2}} = \frac{80}{20} = 4 \quad a_{\mathrm{XFMR1}} = \frac{115}{80} = 1.4375 \]

\[ I_{20V} = 7 \Rightarrow I_{1,\mathrm{XFMR2}} = \frac{7}{4} = 1.75\ \mathrm{A} \]

\[ I_{80V} = 5 + 8 + 4 + 1.75 = 18.75 \Rightarrow I_{1,\mathrm{XFMR1}} = \frac{18.75}{1.4375} = 13\ \mathrm{A} \]

\[ P_{\mathrm{ALT}} = (13)(115) = 1.495\ \mathrm{kW} \]

8. An aircraft requires a 205 \(\mathrm{V}_{RMS}\) bus and a 32 V DC bus. Calculate the transformer turns ratio and breaker values for breakers A and B. The Comm draws 4 A and the Camera draws 5 A. The AC–DC converter efficiency is 80%.

\[ a = \frac{205}{35} = 5.86 \]

\[ P_{\mathrm{Comm}} = (32)(4) = 128\ \mathrm{W} \quad P_{\mathrm{Camera}} = (32)(5) = 160\ \mathrm{W} \]

\[ P_{\mathrm{DC}} = 288\ \mathrm{W} \Rightarrow P_{\mathrm{AC}} = \frac{288}{0.8} = 360\ \mathrm{W} \]

\[ I_{\mathrm{XFMR}} = \frac{360}{205} = 1.75\ \mathrm{A} \Rightarrow \mathrm{CB} \approx 2\ \mathrm{A} \]

9. Is the retrofit design below valid? Why or why not?

No. Transformers only work with AC. Since the added transformer is connected to a DC bus, no voltage or current will appear on the secondary, and the 5 V bus will receive no power.

10. How many breakers should be added to the building security system below, and where?

Solution:

• 8 breakers for individual components

• 2 breakers at the buses

• Up to 3 additional breakers near transformers and the ADC

Total: 10–13 breakers (optimal)