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BartSMP [9]
3 years ago
7

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

running through the entire circuit?
Physics
1 answer:
larisa [96]3 years ago
6 0

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega

And now we can use Ohm's law to find the current in the circuit:

V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A

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I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

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The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

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Which type of electricity is a one-time event, caused by unbalanced charges trying to become neutral again?
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3 years ago
A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.
lyudmila [28]

Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, V_{p} = 120\ V          (rms voltage)

Voltage at secondary, V_{s} = 13000\ V  (rms voltage)

Current in the secondary, I_{s} = 8.50\ mA  

Now,

(a) The ratio of secondary to primary turns is given by the relation:

\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}

where

N_{p} = No. of turns in primary

N_{s} = No. of turns in secondary

\frac{N_{s}}{N_{p}} = \frac{13000}{120} ≈ 108

(b) The power supplied to the line is given by:

Power, P = V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW

(c) The current rating that the fuse should have is given by:

\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}

\frac{13000}{120} = \frac{I_{p}}{8.50}

I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A

 

6 0
3 years ago
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