Let's talk about 4/5.

I'm going to assume you found the frq of 4 and 5 online and as such I will not provide them here. Google em!

4.) THE CIRCUITS FRQ

a.)

Ranking:

(V_a = V_d) > (V_c = V_b)

Reasoning:

A and D take the full current and have the same resistance, so they both have the same voltage drop. C and B take half the current that A and B take but still have the same resistance, so the voltage drop of C vs. D is the same, but compared to A or B, V_c and V_d is less.

b.)

The current through A will DECREASE.

This is because when you take away R_b, you actually RAISE then net resistance.

Notice that R_net=R_a + R_d + R_bc, in which R_bc=1/( 1/(R_b) + 1/(R_c) ). Taking away R_b will take away from the denominator of R_bc, so R_bc increases.

Since the net resistance increases and V_battery is the same, then I_net must decrease by Ohm's law. Since R_a takes the current both before and after the removal of R_b, then I_net=I_a, and lowering I_net means lowering I_a.

c.)

The current through C will INCREASE.

We should probably calculate the current I_c both before and after you remove R_b.

Since all resistors have the same resistance , we can arbitrarily assign this resistance as R.

BEFORE:

R_net=R+R+1/(1/(1/R)+(1/R))=2R+0.5R=5R/2

V_bat=I_net*R_net

V_bat=I_net*5R/2 I_net=(V_bat/R)(2/5)

I_c=I_net/2 because B and C are in parallel with both having the same resistance, so I_c=(V_bat/R)(1/5)

AFTER:

R_net=R+R+R=3R

V_bat=I_netR_net=I_net3R

(V_bat/R)*1/3=I_net

This time, I_net=I_c because C is in series with the battery, so I_c = (V_bat/R)*(1/3)

Notice how I_c increased from 1/5 * V/R to 1/3 * V/R, so then the current will increase.