contestada

A 15-ohm resistor and a 20.-ohm resistor are connected in parallel with a 9.0-volt battery. A
single ammeter is connected to measure the total current of the circuit.

In the space in your answer booklet, draw a diagram of this circuit using symbols from the Reference
Tables for Physical Setting/Physics. [Assume the availability of any number of wires of negligible
resistance.]

Respuesta :

Β  Β  Β The equivalent resistence of the circuit is given by:

[tex]R_{eq}= \frac{R_{1}*R_{2}}{R_{1}+R_{2}} \\ R_{eq}= \frac{15*20}{15+20} \\ R_{eq}= \frac{60}{7} \Omega [/tex]
Β 
Β  Β  Β Using the Ohm's First Law on the whole circuit, we have:

[tex]V=R_{eq}*I_{eq} \\ 9= \frac{60*I_{eq}}{7} \\ \boxed {I_{eq}=1.05A}[/tex]
Β Β 
Β  Β  Β The simulation is attached.

If you notice anyΒ mistake in my english, please let me know, because i am not native.



Ver imagen Аноним

When the resistance are connected in parallel, with the potential difference, the total current of the circuit Β is 1.05 Amperes.

What is resistance?

The resistance in the circuit is the opposition provided to the current flow.

The 15-ohm resistor and a 20 ohm resistor are connected in parallel, their equivalent resistance will be

Req = (R1 x R2) /(R1 + R2)

Substitute the values of resistance, we get

Req = (15 x 20) /(15 +20)

Req = 60/7 ohm.

The total current in the circuit, according to the ohms law is

V =I x Req

Plug the value of potential difference and the equivalent resistance, we get

I = 9 / (60/7)

I = 1.05 Amperes

Thus, the total current in the circuit is 1.05 Amperes

Learn more about resistance.

#SPJ2

Otras preguntas