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#circuit-analysis
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#electrical-engineering
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#transistors
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The voltage drop across a transistor network is 12v if the total resistance is 5.6kohms what is the current level (2)what is the power delivered (3)how much energy is dissipated in 1hour?

Feb 25

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Feb 25

1) To find the current level, we can use Ohm's Law: V = IR, where V is the voltage drop, I is the current, and R is the total resistance.

Given:

V = 12V

R = 5.6kΩ = 5600Ω

Using Ohm's Law:

12 = I * 5600

I = 12 / 5600

I ≈ 0.00214 A

Therefore, the current level is approximately 0.00214 A.

2) To find the power delivered, we can use the formula: P = IV, where P is power, I is current, and V is voltage.

Given:

I = 0.00214 A

V = 12V

Using the formula:

P = 0.00214 * 12

P ≈ 0.0257 W

Therefore, the power delivered is approximately 0.0257 W.

3) To find the energy dissipated in 1 hour, we can use the formula: E = P * t, where E is energy, P is power, and t is time.

Given:

P = 0.0257 W

t = 1 hour = 3600 seconds

Using the formula:

E = 0.0257 * 3600

E ≈ 92.52 J

Therefore, the energy dissipated in 1 hour is approximately 92.52 Joules.

Given:

V = 12V

R = 5.6kΩ = 5600Ω

Using Ohm's Law:

12 = I * 5600

I = 12 / 5600

I ≈ 0.00214 A

Therefore, the current level is approximately 0.00214 A.

2) To find the power delivered, we can use the formula: P = IV, where P is power, I is current, and V is voltage.

Given:

I = 0.00214 A

V = 12V

Using the formula:

P = 0.00214 * 12

P ≈ 0.0257 W

Therefore, the power delivered is approximately 0.0257 W.

3) To find the energy dissipated in 1 hour, we can use the formula: E = P * t, where E is energy, P is power, and t is time.

Given:

P = 0.0257 W

t = 1 hour = 3600 seconds

Using the formula:

E = 0.0257 * 3600

E ≈ 92.52 J

Therefore, the energy dissipated in 1 hour is approximately 92.52 Joules.