Change the value of the battery emf to 10.0 V, and make sure the middle resistor is set to 20 Ω. Use extra wire for each "leg" of the circuit so you can measure the current through all legs with the ammeter. How does the current coming out of the battery change when the switch is closed?

The question is incomplete! The complete question along with answer and explanation is provided below.

Circuit diagram is also attached.

Question:

Construct a circuit having two resistors connected in parallel, as shown in the figure below.

One 20Ω resistor and one 10Ω resistor connected in parallel.

Change the value of the battery emf to 10.0 V, and make sure the middle resistor is set to 20 Ω. Use extra wire for each "leg" of the circuit so you can measure the current through all legs with the ammeter.

How does the current coming out of the battery change when the switch is closed?

Resistors in parallel reduces the equivalent resistance of the circuit which increases the current in the circuit.

Explanation:

According to Ohm's Law

V = IReq

I = V/Req

Resistors connected in parallel reduces the equivalent resistance of the circuit which in turn reduces the current in the circuit.

To understand it better let us first consider a circuit where resistors are connected in series.

Case 1: Resistors in series

R1 = 10Ω and R2 = 20Ω and V = 10V

Req = 10 + 20 = 30Ω

I = V/Req

I = 10/30

I = 1/3 = 0.33 A

Case 2: Resistors in parallel

Now consider the attached circuit where resistors are connected in parallel.

R1 = 10Ω and R2 = 20Ω and V = 10V

Req = R1*R2/R1+R2

Req = 10*20/10+20

Req = 6.67Ω  (note that the equivalent resistance is reduced)

I = V/Req

I = 10/6.67

I = 1.49 A

Hence the current is increased from 0.33A to 1.49A

Conclusion:

Resistors in parallel reduces the equivalent resistance of the circuit.

Resistors in parallel increases the current in the circuit.