Answer:
The current increases when the circuit is closed.
Explanation:
As the complete question is not given, the complete question is attached herewith.
From the data the resistors are connected in parallel and the value of Battery EMF is increased thus as the resistance is decreased and the EMF of the battery is increased, thus the current will increase when the switch is closed.
I₃ = 0.17 A into the junction.
The key to solve this problem is using Kirchhoff's Current Law which statements that the algebraic sum of the currents that enter and leave a particular junction must be 0 (I₁+I₂+...+In = 0). Be careful, the currents that leaves a point is considered positive current, and the one that enters a point is considered negative.
In other words, the sum of the currents that enter to the joint is equal to the sum of the currents that come out of the joint.
Three wire meet at junction. Wire 1 has a current of 0.40 A into the junction, the current of the wire 2 is 0.57 A out of the junction. What is the magnitude of the current in wire 3?
To calculate the current in the wire 3:
First, let's name the currents in the circuit. So:
The current in the wire 1 is I₁ = 0.40 A, the current in the wire 2 is I₂ = 0.57 A, and the current of wire 3 is I₃ = ?.
Second, we make a scheme of the circuit, with the current I₁ of wire 1 into the junction, the current I₂ of wire 2 out of the junction, and let's suppose that the current I₃ of the wire 3 into the junction. (See the image attached)
Using Kirchhoff's Current Law, the sum of the currents I₁ and I₃ into the junction is equal to the current I₂ out of junction.
I₁ + I₃ = I₂
0.40 A + I₃ = 0.57 A
I₃ = 0.57 A - 0.40A
I₃ = 0.17 A
Checking the Kirchhoff's Current Law, the sum of all currents is equal to 0 :
I₁ + I₃ = I₂
I₁ - I₂ + I₃ = 0
0.40 A - 0.57 A + 0.17 A = 0
Let's suppose that the current I₃ of the wire 3 out the junction.
I₁ = I₂ + I₃
0.40 A = 0.57 A + I₃
I₃ = 0.40 A - 0.57 A
I₃ = -0.17 A which means that the current flow in the opposite direction that we selected.
