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DIA [1.3K]
3 years ago
10

An incandescent lightbulb is rated at 120 Watts when plugged into a 200 V-rms household outlet. Calculate the resistance of the

filament and the rms current. g
Physics
1 answer:
valina [46]3 years ago
4 0

Answer:

  • The resistance of the filament is 333.33 ohms
  • The rms current is 0.6 A

Explanation:

Given;

output power of the incandescent lightbulb , P = 120 W

input voltage, V = 200 V

The resistance of the filament is calculated as;

P = \frac{V^2}{R}

Where;

R is the resistance of the filament

R = \frac{V^2}{P} \\\\R = \frac{200^2}{120} \\\\R = 333.33 \ ohms

The rms current is given as;

P = I_{rms} V_{rms}\\\\I_{rms} = \frac{P}{V_{rms}} \\\\I_{rms} =\frac{120}{200}\\\\ I_{rms} = 0.6 \ A

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3 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

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4 years ago
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Answer:

600 Joules

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