Question

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.]

Answer 1

     The equivalent resistence of the circuit is given by:

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

$V=R_{eq}*I_{eq} \\ 9= \frac{60*I_{eq}}{7} \\ \boxed {I_{eq}=1.05A}$
  
     The simulation is attached.

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

Answer 2

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

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