A galvanometer is used as a

In the circuit described by the diagram, which pair of resistors is connected in parallel?

scoundrel [369]

Option 4 ( R2 and R3 ) is the correct answer.

Explanation:

In the below given diagram, we can see a circuit diagram that has four resistors such as R1, R2, R3, and R4.
The opening of the circuit is noted as "a" and the ending is noted as "b".
By observing the above diagram, we can clearly see that R2 and R3 are the pair of resistors that are connected in a parallel manner.
Where all the other resistors such as R1 and R4 are neither connected in parallel nor in series.

Hence we can conclude that Resistor R2 and R3 are the ones that are connected in parallel.

6 0

4 years ago

I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!

sergij07 [2.7K]

Answer:

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

V = 12 V
P = 24 W

$\implies \mathsf{24=\frac{12^2}{R} }$

$\implies \mathsf{24R=12^2 }$

$\implies \mathsf{24R=144 }$

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

<em>R and 100 can interchange places</em>

$\implies \mathsf{R=\frac{240^2}{100} }$

$\implies \mathsf{R=\frac{57600}{100} }$

<u></u>

By Ohm's Law:

$\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}$

=> 240 = I × 576

=>

=> I = 0.417 A

I don't know it's resistance,... so sorry

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

3 0

3 years ago

Please help with this

sertanlavr [38]

Answer:I have to say 56

Explanation: because it is going up by 8

6 0

3 years ago

A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp

Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0

2 years ago

Define total mechanical energy

zavuch27 [327]

Answer:

The total mechanical energy is the sum of kinetic and potential energies: E = K + U. The law of conservation of total mechanical energy states that the sum of the kinetic energy and potential energy is constant in time.

3 0

3 years ago