Can someone please answer and explain why you would use 30A?? E = 120 v P_2 = V^

Question

Can someone please answer and explain why you would use 30A??

E = 120 v P_2 = V^2/R_2 rightarrow R_2 = V^2/P_2 R_ 2 = (120 V)^2/75 W = 19256 V_1 = IR_1 = (3 o A) (0.456) = 12 V V_P = V- V_1 = 120 - 12 V = 108 V P_2 = VP^2/R_2 = (108V)^2/19256 = 60.75 W b)P_3 = P_sat - P_2 = IV - P_2 - P_1 = (3 o A) (1 L o V) - (60.75w) - (30A)^2 (0.456) = 3.18 times 10^3 W Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is 0.400 Ohm, and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance, (b) What power is consumed by the motor? Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant IR drop in the wires and reduces the voltage across the light.

Explanation / Answer

because the resistance R2 and R3 are parallel cinnection

the resistance R2 and R3 are equal so the current i2=i3=15 A

step;2

now we find the power on resistance R2

power P=120^/75=192 ohm W

step;3

now we find the voltage on resistance R1

the current v=v=30*0.4=12 v

the voltage on parallel connection vnet=120-12=108 v

step;4

now we find the power at resistance in resistance a

power P=30*120-

P2=108^2/12=530.2 w

(b)m ans

power P=30*120 v--50.8*1^-velocity at hera

= **1oj

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