All categories

2 years ago

As more lamps are put into a series circuit, the overall current in the power sourceA) increases.B) decreases.C) stays the same.

1 answer:

8 0

As more and more lamps are connected in parallel (and if the current does not produce heating inside the battery) their brightness stays the same. Each lamp has the same voltage across it. Each lamp added in parallel decreases the total resistance in the circuit, so additional current flows.

You might be interested in

75 pts!

jasenka [17]

again CORRECT ans is 4) light strikes the grooves → different wavelengths of light bend at different angles → diffracted wavelengths reach the eyes → the eyes see different colors.

moderator - plz review the ans as u deleted my right ans n approved the wrong ans :(

6 0

2 years ago

Read 2 more answers

Sound is a disturbance that travels through a medium as a

Vinvika [58]

Answer:

Sound waves. Anything that vibrates is producing sound; soundis simply a longitudinal wave passing through a medium via the vibration of particles in themedium. Consider a sound wavetraveling in air

5 0

3 years ago

a DJ uses a 110-volt outlet to plug in a strobe light. if the current flowing through the light is 0.050 amps, how much resistan

tigry1 [53]

Ohms Law: V = IR

V is the voltage in volts

I is the current in amps

R is the resistance in Ohms

Rearrange: R = V/I

R = (110)/(0.050)

R = 2200

There are 2200 Ohms of resistance in the circuit.

5 0

3 years ago

A 65-kg person walks from the ground to the roof of a 100 m tall building. How much gravitational potential energy does she have

ivolga24 [154]

I,think potential energy is mgh so 65*100*9,81

7 0

2 years ago

A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

Where

L = Inductance

C = Capacitance

Our values are given as

$C = 15*10^{-6}\mu F$

$L = 0.280*10^{-3}mH$

Replacing we have,

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})$

$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0

3 years ago