A 240-volt, 2-amp motor is connected to a three-wire, 120/240-volt system. Connected between the black wire and neutral are four 200-watt, 120-volt lamps and a 120-volt, 1-amp motor. Between the red wire and neutral are three 200-watt, 120-volt lamps, one 1.67-amp motor and one 120-volt, 1-amp motor. (Round the FINAL answer to two decimal places.). How many amps flow in the red wire?

(i)The current flow in black wire = 9.67 A (ii) The current low in the red wire is 9.68 A (iii) The current flow in neutral wire is  15.36 A (iv) when 240 volt were disconnected current  in black wire is 7.68 A (v) when 240 volt were disconnected current in red wire is 7.68 A (vi) 15.36 A (vii) 6.34 (viii) 9.68 A (ix) 12.02 A

Explanation:

Solution

The current drawn by one amp is

I =P/V

I =200/120

I= 1.67 A

(i) The current flow in the black wire  is

IBK = 4 * 1.67 A + 1A + 2A

IBK = 9.67 A

(ii) Current flow in the red wire is

IRD = 3 * 1.67 A + 1.67 A + 1A + 2A

= 8.68A + 1 A = 9.68 A

Note: Kindly find an attached copy of part of the solution to the given question above.

i believe the answer would be:

d. violin, clarinet, french horn, and tuba