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3 years ago

Help me besties PLEASE ;-;

Physics

1 answer:

Svet_ta [14]3 years ago

6 0

Answer:

Current in resistors = 1.5 amp

Current in resistors = 0.18 amp

P = 93.75 watts

Explanation:

Given:

1. Three 5 ohm resistor(series)

Voltage = 10 volt

2. Three 5 ohm resistor(parallel)

Voltage = 9 volt

3. Three 5 ohm resistor(series)

Voltage = 6 volt

Find:

Current in resistors

Total power of circuit

Computation:

1. Total resistor (Series) R = R1 + R2 + R3

Total resistor (Series) R = 5 + 5 + 5

Total resistor (Series) R = 15 ohm

I = V/R

Current in resistors = (15)/(10)

Current in resistors = 1.5 amp

2. Total resistor (Series) R = 1/[1/R1 + 1/R2 + 1/R3

Total resistor (Series) R = 1[1/5 + 1/5 + 1/5]

Total resistor (Series) R = 1.67 ohm

I = V/R

Current in resistors = (1.67)/(9)

Current in resistors = 0.18 amp

3. Total resistor (Series) R = R1 + R2 + R3

Total resistor (Series) R = 5 + 5 + 5

Total resistor (Series) R = 15 ohm

I = V/R

I = (15)/(6)

I = 2.5 amp

P = I²(R)

P = (2.5)²(15)

P = 93.75 watts

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A. 0.77 A

Using the relationship:

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

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

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$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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3 years ago

Help me besties PLEASE ;-;​

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