All categories

3 years ago

a resistance of 30 ohms is placed in a circuit with 90 volt battery what current flow in the circuit

Physics

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

3amps

Explanation:

voltage v = current I × resistance R

current = voltage÷ resistance = 90÷30= 3amperes

You might be interested in

Explain what happens to the pitch of a cell phone ring when the amplitude of a sound wave increases. Justify your reasoning usin

dalvyx [7]

<span>Nothing happens to the pitch of a cell phone ring when the amplitude
of a sound wave increases.

Pitch and amplitude are both characteristics of a wave, but they're not
connected, and they don't influence each other.</span>

7 0

3 years ago

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

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.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

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

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.

7 0

3 years ago

How much energy does a 100-W light bulb use in 10 s?

telo118 [61]

Answer:c

Explanation:

8 0

3 years ago

We cannot consider something to be work unless

iren [92.7K]

<span>force applied causes movement of an object in the same direction as the applied force.</span>

5 0