The concept behind this is that the voltage drop across each branch in a parallel circuit is the same. But current is different among the branches. Take the two situations below. Just by closing the switch, the voltage of the battery does not change 100volts must be supplied across each branch initially. If this is true the current through R1 both before and after, must stay the same because the resistance of R1 does not change. V=IR, V(same before and after)/R(same before and after) = I(same before and after).
What does change, however, is the current through the middle branch as can be seen below: For clarity sake, you can try this math approach below just so you understand it, but definitely will waste time on test day....try to understand the above concept through the math. Hope this helps.
Voltage of battery=100v, each resistor = 5 ohms
Before:
Top branch: Voltage across first resistor R1 = 100v, R1=5 ohms, R2, 5ohms, Itop=100Vtotal/10ohms=10A
Bottom branch: Voltage across first resistor R3=100v, R3=5ohms R4=5ohms, Ibottom=100Vtotal/10ohms=10A
Req= (10*10)/(10+10)=100/20=5ohms
I total= 100V/5= 20A=Itop+Ibottom
After:
Top branch: Voltage across first resistor R1= 100v, R1=5 ohms, R2=5ohms, 100Vtotal/10ohms=10A, Itop=100Vtotal/10ohms=10A
Middle branch: Voltage across resistor R5= 100V, R5=5ohms, 100V/5ohms=20A, Imiddle=100Vtotal/5ohms=20A
Bottom branch: Voltage across branch R3= 100v, R3=5 ohms, R4, 5ohms, 100Vtotal/10ohms=10A, Ibottom=100Vtotal/10ohms=10A
Req=(5*5)/(5+5)=2.5ohms=Itop+Imiddle+Ibottom
Votal/Req= 100V/2.5= 40A
Thus, the current across branches 1 and 3 do not change, but when you add the middle branch in, the total equivalent resistance decreases, which leads to an increase in the total current, manifested through an increase in 20A being supplied only to branch 2, I1 and I3 stay constant.
Cheers.