find voltage and current?

Q1:
Refer to Figure 1 below.

Equivalent resistance of R₁ and R₃, R₁₃ = 6.8 + 3.3 = 10.1 kΩ
Equivalent resistance of R₂ and R₄, R₂₄ = 1.5 + 15 = 16.5 kΩ
Equivalent resistance of the 4 resistors = 1/[(1/10.1) + (1/16.5)] = 6.27 kΩ

Voltage across the main circuit, V = iR = 13.5 × 6.27 = 84.6 V

Current passing through A, i₁ = V/R₁₃ = 84.6/10.1 = 8.38 mA
Voltage between C and A = Vca = i₁R₁ = 8.38 × 6.8 = 57.0 V

Current passing through B, i₂ = V/R₂₄ = 84.6/16.5 = 5.13 mA
Voltage between C and B = Vcb = i₂R₂ = 5.13 × 1.5 = 7.7 V

Voltage between A and B = 57.0 - 7.7 = 49.3 V

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Q2:
Refer to Figure 2 below.

Current flows through A and B = 13.5 mA

a)  With A and B unconnected

RA = R1 in series with R3 = 6.8+3.3 = 10.1 kΩ
RB = R2 in series with R4 = 1.5 + 15 = 16.5 kΩ
Rtot = RA in parallel with RB = RARB/(RA+RB) = 10.1*16.5/(10.1+16.5) = 6.265 kΩ
Voltage from supply = I.Rtot = 13.5mA*6.265kΩ = 84.6V

VA = 84.6*6.8/(6.8+3.3) = 56.96V (potential divider formula)
VB = 84.3*1.5/(1.5+15) = 7.66V
VAB = 56.96 – 7.66 = 49.3V
_____________________

b) With A and B connected

RC = R1 in parallel with R2 = 6.8*1.5/(6.8+1.5) = 1.229kΩ
RD = R3 in parallel with R4 = 3.3*15/(3.3+15) = 2.705kΩ
Rtot = 1.229+2.705= 3.934kΩ
Voltage from supply = I.Rtot = 13.5mA*3.934kΩ = 53.1V

Using potential divider formula, voltage across RC = 53.1*(1.228/3.934) = 16.58V
This is the same as the voltage across R1 (and R2).
Current through R1 (=V/R) = 16.58/6.8 = 2.44mA

Voltage across R3 (and R4) = 53.1 – 16.58 = 36.52V
Current through R3 (=V/R) = 36.52/3.3 = 11.07mA

Consider junction between R1 and R3.  Current flows from B to A (since the current through R3 is bigger than the current through R1).

Current from B to A is the difference in currents between currents through R1 and R3
IAB =  11.07 – 2.44 = 8.63mA = 8.6mA rounded.

Q:1) R1& R3 in series & R2 & R4 are in series, let their resultant resistence is S & T respectively,
=>By R(net) = R1 + R2 + R3 + -------- (for series)
=>S = R1 + R3  = 6.8 + 3.3 = 10.1 kΩ
& T = R2 + R4 = 1.5 + 15 = 16.5 kΩ
Now S & T are in parallel, let their resultant resistence is N, 
=>By 1/R(net) = 1/r1 + 1/R2 + 1/R3 + ------(for parallel)
=>1/N = 1/S + 1/T = 1/10.1 + 1/16.5 
=>N = 6.26 kΩ
Thus Voltage across S & T will be, by V = iR
=>V = iN
=>V = 13.5 x 10^-3 x 6.26 x 10^3
=>V = 84.51 V
Let the current through R1 & R2 is i1 and i2 through R2 & R3
=>By V = iR
=>V = i1S 
=>84.51 = i1 x 10.1 x 10^3
=>i1 = 8.37 mA
Thus by i2 = i - i1 
=>i2 = 13.5 - 8.37 = 5.13 mA
Now Voltage drop across R1 by V = iR
=>V(R1) = i1 x R1 = 8.37 x 10^-3 x 6.8 x 10^3 = 56.92 V
Thus Voltage at A i.e. V(A) = V - V(R1) = 84.51 - 56.92 = 27.59 V
Now Voltage drop across R2 by V = iR
=>V(R1) = i2 x R2 = 5.13 x 10^-3 x 1.5 x 10^3 = 7.70 V
Thus Voltage at B i.e.V(B) = V - V(R2) = 84.51 - 7.70 = 76.81 V
Thus voltage between Points A and B = |V(A) - V(B)| = 49.22 V
Q:2) As A & B are connected
=>R1 & R2 are in parallel and R3 & R4 are in parallel now, let their resultant resistance is S & T respectively,
=>1/S = 1/6.8 + 1/1.5 
=>S = 1.23 kΩ
& 1/T = 1/3.3 + 1/15
=>T = 2.70 kΩ
Now S & T are in series, Let their resultant resistance is N, 
=>N = S + T
=>N = 1.23 + 2.70
=>N = 3.93 kΩ
Thus current through will be equal in S & T i.e. 13.5 mA, however ir will be different in R1, R2, R3 & R4. Let it is i1, i2, i3 & i4 respectively,
=>i1R1 = i2R2 = i x S
=>i1 R1 = i x S
=>i1 = (13.5 x 10^-3 x 1.23 x 10^3)/(6.8 x 10^3) = 2.44 mA
& i2 R2 = i x S
=>i2 = (13.5 x 10^-3 x 1.23 x 10^3)/(1.5 x 10^3) = 11.06 mA
& =>i3R3 = i4R4 = i x T
=>i3 R3 = i x T
=>i3 = (13.5 x 10^-3 x 2.70 x 10^3)/(3.3 x 10^3) = 11.06 mA
& i4 R4 = i x T
=>i2 = (13.5 x 10^-3 x 2.70 x 10^3)/(15 x 10^3) = 2.44 mA