. Class Management I Help Hw Assignment #23 Begin Date: 7/16/2018 1:00:00 PM ..

Question

. Class Management I Help Hw Assignment #23 Begin Date: 7/16/2018 1:00:00 PM .. Due Date: 7 (33%) Problem 1: Answer the following about the figure. 18/2018 i :00:00 PM End Date: 7/27/2018 1:00:00 PM 90 2 10 2 10 V 452 40 22 pleted l pleted | ? 33% Part (a) what is the current (in A) supplied by the battery to the electric circuit? e 33% Part (b) what is the power (in mW) dissipated by the 45 #13% Part (c) What is the power (in mW) dissipated by the 90 a resistor? | | resistor? Grade Summary 415 0% 100% Potential sin() | cos() | tan() | ?? ( D 7 | 8| 9|HOME cotan asin0 atan acotan sinh0 coshO tanh cotanh0 Attempts remaining: 17 (0%per attempt) detailed view acoso 0% 0% 0% 23 ODegrees Radians BACKSPACE RL CLEAR Submit l give up Hints: 0 for a 0% deductionHiatsfernaning: 0 Feedback: 0% deduction per feedback

Explanation / Answer

(a) 90ohm and 45 ohm in parallel, their equivalent = (90 x 45)/(90+45) = 30 ohms

then this 30 ohms and 10 ohm and 40 ohm in series, gives REq = 30+10+40 = 80 ohms

therefore, current = 10V / 80 ohms = 0.125 A Answer

(b) Current in 45 ohms = 0.125 x 90/(45+90) = 0.083 A

Therefore, Power dissipated in 45 ohms = 0.0832 x 45 = 312.5 mW Answer

(c)

Current in 90 ohms = 0.125 x 45/(45+90) = 0.0417 A

Therefore, Power dissipated in 90 ohms = 0.04172 x 90 = 156.25 mW Answer

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