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Romashka-Z-Leto [24]

3 years ago

6

Two identical bulbs in parallel in a radio create a total resistance of 15 ohms in the circuit. What's the resistance of each of

the bulbs A. 10 ohms B. 7.5 ohms C. 45 ohms D. 30 ohms

1 answer:

Tatiana [17]3 years ago

5 0

Answer

letter B is the answer because 7.5+7.5=15

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Draw the logic circuit for each of the following. For each gate, determine if it generates either EVEN or ODD parity bit and fin

omeli [17]

Answer:

a) 4-input XOR, input data-1001 = 0 Even parity Bit

b) 5-input XOR, input data-10010 = 0 Even parity Bit

c) 6-input XOR, input data-101001 = 1 Even parity Bit

d) 7-input XOR, input data 1011011 = 1 Even parity Bit

Explanation:

a) 4-input XOR, input data-1001 ; generates 0 Even parity Bit

b) 5-input XOR, input data-10010 ; generates 0 Even parity Bit

c) 6-input XOR, input data-101001 ; generates 1 Even parity Bit

d) 7-input XOR, input data 1011011 ; generates 1 Even parity Bit

Attached below is the Logic circuits of the data inputs

8 0

3 years ago

Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The eq

evablogger [386]

Answer:

A) attached below

B) I₁ = 18.1 A , I₂ = 69.39 A

C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%

Explanation:

A) Schematic diagram attached below

attached below

<u>B) magnitude of primary and secondary winding currents </u>

I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )

VL = Line voltage = 208

cos∅ ( power factor ) = 0.8

P = 20 * 10^3 watts

insert values into equation 1

I₂ = 69.39 A

I₁ ( primary current ) = I₂V2 / V1

I₁ = ( 69.39 * 120 ) / 460 = 18.1 A

<u>C ) Calculate the Primary voltage magnitude and the Voltage regulation</u>

V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )

= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )

= 460 + 40.544 ∠ 100.3°

∴ V( magnitude) = 454.5 ∠ 5.04° V

<em>Voltage regulation </em>

= ((Vmag - V1) / V1 )) * 100

= (( 454.5 - 460 / 460 )) * 100

= -1.195 % ≈ -1.2%

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3 years ago

Why is the drawdown cone for a well completed in a low permeability aquifer narrower and deeper than a drawdown cone for a well

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Answer:

c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer

Explanation:

The groundwater are contains under the rock and in the open spaces within the rocks and the unconsolidated sediments. Aquifer refers to the underground layers of the permeable sand or rocks that transmits the groundwater below water table which provides a sufficient supply of water to the well. Groundwater is present everywhere where there is porosity in the rocks and it depends on the permeability of the rocks to allow them flow.

A drawdown cone is completed in the lower permeable aquifer deeper and narrower than the high permeable aquifer as it takes more amount hydraulic head or energy to drive groundwater to the well casing which is in the lower permeable aquifer.

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Fact

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Answer:

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