Question

AMA INTERNATIONAL UNIVERSITY-BAHRAIN Salmabad, Kingdom of Bahrain COLLEGE OF ENGINEERING 1st Tri, 2nd Tri 3rd Tri Summer, SY 2014-2015 Prelim Exam ENGG03D/ENGG531 Electric Circuit TI II PART A: Kirchhoff s Voltage Law Using only KVL find the value of vo e 2v 150V V PART B: Series Parallel circuit (15) Determine the value of Rob in the given circuit below: 300 400 200 600 100 500 40 700 800

Answer 1

Using KVL in loop ABCDA , we get the equation as :

(20 ohm) I + (-2V_o) + (25 ohm) I + (-V_o) + (-150) =0

20I - 2V_o + 25I - V_o - 150 =0              Eq-1

In branch CD of the circuit , Voltage = V_o , current = I and resistance = 20

so using ohm's law , V_o = 20I

replacing ''V_o'' by ''20I'' in Eq-1 ,

20I - 2V_o + 25I - V_o - 150 =0      

20I - 2(20I) + 25I - 20I - 150 =0

20I - 40I + 25I - 20I -150 =0

-15I -150 =

-15I = 150

I = -10 A

so V_o = 20 I = 20 (-10) = -200 Volts             (since I = -10 A)

B)

     
Зол point Q and S cań b joined 10 ohm and 50 ohm are in series sinde both are joined only at one 1on S0 SL 1 o -( DA个 ohm are in series since both are joined only at one end to each other

series Combination of resistance = R1 + R2

so Combination of 10 ohm and 50 ohm = 10 + 50 = 60 ohm,

similarly , combination of 70 oh and 80 ohm = 70 + 80 = 150 ohm

Diagram Now looks as

フ 3L 30 ohm and 20 ohm are in parallel since they are connected to each other at both end 51 15o1

parallel Combination of reistances = $\frac{R_{1}\times R_{2}}{R_{1} +R_{2}}$     ,   

So parallel combination of 30 ohm and 20 ohm = $\frac{30\times 20}{30 +20}$ = 12 ohm

parallel combination of 40 ohm and 60 ohm = $\frac{40\times 60}{40 +60}$ = 24 ohm

parallel combination of 60 ohm and 150 ohm = $\frac{60\times 150}{60 +150}$ = 42.86 ohm

so Diagram now looks as

all resistances are now in series.

so total resistance is simply the sum of all.

so R_total = 8 ohm + 12 ohm + 24ohm + 6 ohm + 42.86 ohm + 4 ohm = 96.86 ohm