Solved Examples of Comlicated Circuits

Illustration:

Let us analyse a simple circuit shown in the figure alongside. Assume current values (I₁, I₂ & I₃) at random directions.  

Alt txt:  simple circuit     

Solutions                     

Þ     All through the branch gfdab current in I₁

        All through the branch geb current is I₃

        All through the branch ghcb current is I₂

        Applying KCL at b Þ I₁ + I₂ + I₃ = 0                       …… (i)

        (Note: We will get the same eqn. at node g)

The voltage drops across the circuit elements in loop fdbegf,

        V_d – V_f = 10V              (constant voltage source)

        V_d – V_a = 1 × I₁           (drop in the direction of current flow)

        V_e – V_b = 2 × I₃           (drop in the direction of current flow)

        V_g – V_e = 15V             (constant voltage source)

Moving anti clockwise we write,

KVL: (V_f – V_d)+(V_d – V_a)+(V_a – V_b)+(V_b – V_e)+(V_e – V_g)+(V_g – V_f) = 0

        Þ (–10) + (I₁) + (0) + (–2I₃) + (–15) + 0 = 0

        Þ I₁ – 2_I3 = 25                                                   …… (ii)

Consider loop bchgeb. The voltage drop across the elements is,

        V_h – V_c = 4I2

        V_e – V_b = 2I3

        V_g – V_e = 15

KVL: (V_b – V_c) + (V_c – V_h) + (V_h – V_g) + (V_g – V_e) + (V_e – V_b) = 0

        Þ (0) + (–4I₂) + (0) + (15) + 2I₃ = 0

        Þ 2I₃ – 4I₂ = –15                                                …… (iii)

Now solve (i), (ii), & (iv) simultaneously, you shall arrive at,

        I₁ = 120/14, I₂ = 5 – 5/14, I₃ = –115/14  

Illustration:  

What is the potential difference between the points M and N for the circuits shown in the figures, for case l and case ll?  

                             Alt txt : Potential-difference-between-points

Solution:  

Case l:

                l = E₁ – E₂ / r₂ + r₁ = 12 – 6 / 3 + 2 = 1.2 A

Alt txt: circuit-1

        For cell E₁: v_A – E₁ + lr₁ = v_B

        i.e. v_A – v_B = E₁ – lr₁

                        = 12 – 1.2 × 3 = 8.4 V

        For cell E₂, v_C – E₂ – lr₂ = v_D

        i.e. v_C – v_D = 6 + 1.2 × 2 = 8.4 V

        Hence, v_C – v_D = v_A – v_B = v_M – v_N = 8.4 V  

Case ll:

                l = E₁ + E₂ / r₁ + r₂

                  = 12 + 6 / 3 + 2 = 3.6 A

Alt txt: Circuit-2

        For cell E₁:

                v_A – E₁ + lr₁ = v_B,

        i.e.    v_A – v_B = E₁ – lr₁ = 12 – 3.6 × 3 = 1.2 V

        For cell E₂:

                v_C + E₂ – lr₂ = v_D

                i.e. v_C – v_D = –E₂ + lr₂ = – 6 + 3.6 × 2 = 1.2 V

        Hence, v_A – v_B = v_C – v_D = v_M – v_N = 1.2 V

Illustration:  

In the adjacent circuit, find the effective resistance between the points A and B.

Alt txt: effective resistance between two points

Solution:  

Resistors AF and FE are in series with each other. Therefore, network AEF reduces to a parallel combination of two resistors of 6 W each.

        R_eq = 6 × 6 / 6 + 6 = 3 W.

Similarly, the resistance between A and D is given by 6 × 6 / 6 + 6 = 3 W.

Now, resistor AC is in parallel with the series combination of AD and DC. Therefore, resistance between A and C is 6 × 6 / 6 + 6 = 3 W. Now, the combination of (AC + CB) is in parallel to AB. Therefore, since AC and CB are in series, their combined resistance = 3 + 3 = 6 W. Resistance between A and B is given by, 1 / R = 1/6 + 1/3 = 3/6 or R_AB = 2W.

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