Q.1.a) State Kirchhoff’s laws with suitable example. [05 M]

**Ans. **

There are two Kirchoff’s laws and these laws are used to determine the following –

i) the equivalent resistance of a complicated network of conductors and

ii) for calculating the current flowing in various conductors.

These two laws are stated below:-

i) Kirchoff’s Current Law (KCL) :- It states that ”In any electrical network, the algebraic sum of the currents meeting at a point (or junction) is zero. ”

Simply, In another way it means that total current leaving a junction is equal to total current entering that junction.

Example- Consider few conductors meeting at a point A as shown in fig. We can see that some current flowing towards the junction and some are flowing away from the junction.

Let’s take incoming current positive and outgoing current to be negative .

I_{1} + (-I_{2}) +(-I_{3}) + (+I_{4}) + (-I_{5}) = 0

I_{1}+I_{4}-I_{2}-I_{3}-I_{5}=0

I_{1}+I_{4 }= I_{2} + I_{3} + I_{5}

Incoming current = Outgoing current

This was KIrchoff’s Current law.

ii) Kirchoff’s Voltage Law (KVL) :- It states that ” The algebraic sum of product of current and resistance in each conductors in any closed path in a network plus the algebraic sum of the emfs in that path is zero. ”

In other words,

\sum IR + \sum e.m.f =0Here, the algebraic sum is sum that takes into account the polarities of the voltage drops. This law basis is that if we start from the particular junction and go round the mesh till end come back t starting point, then must be at same potential with which we started.

Ex- Consider the circuit below, ABCDA . Now applying KVL we have,

E-V_{1}-V_{2}-V_{3}=0

E-IR_{1}-IR_{2}-IR_{3}=0

IR_{1}+IR_{2}+IR_{3}=E

Sum of voltage drops= Sum of E.M.Fs or voltage rises

KVL can be applied to any closed circuit.

Q.1. b) Find the voltage across branch AB using super- position theorem.

**Ans. **

**Given :-**

**To Find :- ** Voltage across branch AB V_{AB}=?

**Solution** –

Since there are two sources , let

V=V_{1}+V_{2}

Where, V_{1 }and V_{2} are contributions due to 20 V voltage source and 8A current source respectively.

i) For obtaining V_{1}, deactivating the current source .

Applying KVL to the loop

9I_{1}-20=0

9I_{1}=20

I_{1}=20/9 = 2.22 A

**I _{1}= 2.22 A **

Thus, V_{1} = 4i_{1 }= 4 x 2.22 = 8.88 V

**V _{1}=8.88 V**

Now for calculating V_{2 }, deactivating the second source of the circuit.

i_{3}=\frac{5}{4+5}= \frac{5}{9}= 0.55 A

V_{2}=4i_{3} = 4 x 0.55 = 2.22 V

I_{AB}= I_{1} – I_{3}= 2.22-0.55 = 1.67

V_{AB}= I_{AB} x 4 = 1.67 x 4 = 6.68 V

**V _{AB}= 6.68 V **

The voltage across A and B is 6.68 V .

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