Kirchhoff's Laws: Kirchhoff's Current and Voltage Laws & Applications

Ans. Given that P = 630 W and V = 210 V. 

In DC source, P = VI.

Therefore, I = P/V = 630/210 = 3 A.

Ans. Applying Kirchhoff’s second rule in the closed mesh ABFEA

Ans. By applying Kirchhoff’s rule, we will get 200-10= 190 V

and I = V/R = 190/38 = 5 A

Ans. Applying Kirchhoff’s law by moving along ACDE, we will get

Ans. By applying the loop rule to the network PQRP,

(- 4) = 60 (I - I₁) - 20 I₁ = 0

Or, (- 4) = 60I - 60I₁ - 20I₁

Or, 20I₁ - 15I = 1 [+ by 4… (i)]

Applying loop yule to loop PRSP, we will get

(-5) +200 I + 20 I₁ = 0

4I₁ + 40 I = 1 [+ by 5… (ii)]

Hence, reading of milliammeter is 0.064A

Ans: KVL: Kirchhoff’s voltage law, states that the sum of electromotive forces in a loop equals the sum of potential drops in the loop. Alternatively, it could be stated that the directed sum of voltages around any closed loop is zero.

KCL: Kirchhoff’s current law, also known as Kirchhoff’s junction rule, states that the algebraic sum of currents in a network of conductors meeting at a point is zero.

The above laws are justified because of the law of conservation of energy and law of conservation of charge. 

Ans.If we apply Kirchhoff’s current law in the given circuit

At junction B,

i₁ = i_g + i₃

At junction D,

i₂ + i_g = i₄

If current through the galvanometer is zero,

i_g = 0

Thus, i₁ = i₃ and i₂ = i₄

Applying Kirchhoff’s voltage law for loop ABDA,

i₁P + i_gG = i₂R

Applying Kirchhoff’s voltage law for loop BCDB,

i₃Q = i₄S + i_gG

When i_g = 0, i₁P = i₂R and i₃Q = i₄S

But i₁ = i₃ and i₂ = i₄

Hence, P/Q = R/S.

Ans. By using Kirchhoff’s first law at junction E, we get

I₃ = I₃ + I₂

Using Kirchhoff’s second law in the loop ABCDA

80 - 20 I₂ + 30 I₁ = 0

2 I₂ – 3 I₁ = 8     ( divide by 10)….(ii)

In loop ABFEA, we get

80 – 20 I₂ +20 – 20 I₃ = 0

I₂ + I₃ = 5        (divide by 20)

Putting the value of I3 into (iii), we have

I₂ + (I₁ + I₂) = 5 = 2I₂ + I₁ = 5    ….(iv)

Solving equations (ii) and (iv), we get

I₁ = – ¾ A = – 0.75 A

Therefore, (-) sign of current indicates that the direction of current is opposite to that as shown in the figure.

Ans. The Kirchhoff’s rules are,

i) Junction rule or current rule: The sum of all current entering a junction is equal to sum of all currents leaving the junction.

ii) Voltage rule: The algebraic sum of changes in the potential around any closed loop must be zero.

By applying Kirchhoff’s voltage rule to the loop ABCDA, we get

80 – 20i₁ – 40(i₁ – i₂) = 0

80 – 60i₁ + 40i₂ = 0

4 – 3i₁ + 2i₂ = 0         ….(i)

By applying Kirchhoff’s voltage rule 2 to the loop FEDCF, we get

40 + 40(i₁ – i₂) – 10i₂ = 0

40 + 40i₁ –  50i₂ = 0

4 + 4i₁ + 5i₂ = 0         ….(ii)

Now, multiplying equation (1) by 4 and equation (2) by 3 and then adding them, we get

4(4 – 3i₁ + 2i₂) + 3(4 – 4i₁ + 5i₂) = 0

16 – 12i₁ + 8i₂ + 12 + 12i₂ – 15i₂ = 0

28 – 7i₂ = 0

28 = 7i₂ 

i₂ = 4 A

Substituting i₂ = 4 in equation 1, we get

4 – 3i₁ + 2 x 4 = 0

4 – 3i₁ + 8 = 0

I₂ – 3i₁ = 0

i₁ = 4A

Hence, the current passing through 20Ω resistor is 4 A and the current passing through 40Ω resistor is, i₁ - i₂ = 4 A - 4 A = 0 A

Ans. By considering the loop ABDA

Apply KVL to ABDA,

1 – 2 – 2I₂ – 2(I₁ + I₂) – 1 x I₂ = 0

 – 2I₁ – 5I₂ = 1…..(i)

Apply KVL to DCBD,

3 – 3I₁ – 1 x I – 1 – 2(I₁ + I₂) = 0

 – 6I₁ – 2I₂ = – 2

6I₁ + 2I₂ = 2   …....(ii)

By solving equation 1 and 2, multiply equation (1) by 3

 – 6I₁ – 15I₂ = 3 …...(iii)

Add equation (3) and(2)

 – 13I₂ = 5

I₂ = – 5/13

From equation (1)

 – 2I₁ – 5 ( – 5/13) = 1

 – 2I₁ = 1 – 25/13

 – 2I₁ = – 12/13

 I₁ = 6/13

Hence, current passing through the branch BD,

= I₁ + I₂

= 1/13 A

Magnitude voltage across BD,

= I/13 x 2 = 0.153 volt

Ans. According to Kirchhoff’s rules,

• At any junction, the sum of all current entering a junction is equal to the sum of all currents leaving the junction.
• The algebraic sum of changes in the potential around any closed loop that involves resistors and cells must be zero.

The circuit diagram of a Wheatstone Bridge is given below,

P, Q, R and S are four resistances that are forming a closed bridge known as Wheatstone bridge. A battery is connected between A and C, on the other hand a galvanometer is connected between B and D. There is no current in the galvanometer at balance. 

Now, let the current given by the battery in the balanced position be I. This current after reaching point A, is divided into two parts, I₁ and I₂. As there is a lack of current in the galvanometer in the balanced state, current in resistances P and Q is I₁ and it is I₂ in resistances R and S.

Applying Kirchhoff’s I Law at point A

I – I₁ – I₂ = 0 or I = I₁ + I₂…...(i)

Applying Kirchhoff’s II Law to closed mesh ABDA

 – I₁P + I₂R = 0 or I₁P + I₂R ...(ii)

Applying Kirchhoff’s II Law to closed mesh BCDB

 – I₁Q + I₂S = 0 or I₁Q + I₂S ...(iii)

Dividing equation (ii) by (iii), we get

I₁P/I₁Q = I₂R/I₂S or P/Q = R/S ….(iv)

This shows the condition of the balance of the Wheatstone bridge.

Ans. Following are the things that you should keep in mind while approaching the problem:

• You have to choose the direction of the current. Let us choose the clockwise direction in this problem.
• There is a potential decrease when the current flows through the resistor. As a result, V = IR is signed negative.
• If the current moves from low to high then the source of emf (E) signs positive because of the charging of energy at the emf source. Likewise, if the current moves from high to low voltage (+ to -) then the source of emf (E) signs negative because of the emptying of energy at the emf source.

In this solution, the direction of the current is the same as the direction of clockwise rotation.
(- IR₁+ E₁ - IR₂ - IR₃ - E₂ )= 0
Substituting the values in the equation, we get
(-2I + 10 - 4I - 6I - 5) = 0
(-12I + 5) = 0
I = (-5/-12)
I = 0.416 A

The electric current that flows in the circuit is 0.416 A. The electric current is signed positive which means that the direction of the electric current is the same as the direction of clockwise rotation. If the electric current is negative then the direction of the current would be in an anti-clockwise direction.

Ans. Assume currents to flow in directions indicated by arrows.

Apply KCL on Junctions C and A.

Therefore, current in mesh ABC = i₁

Current in Mesh CA = i₂

Then current in Mesh CDA = i₁ - i₂

Now, applying KVL on Mesh ABC, 20V is acting in a clockwise direction. Equating the sum of IR products, we get;

10i₁ + 4i₂ = 20 ……………. (1)

In mesh ACD, 12 volts are acting in clockwise direction, then:

8(i₁-i₂) - 4i₂= 12

8i₁ - 8i₂ - 4i₂= 12

8i₁ - 12i₂ = 12 ……………. (2)

Multiplying equation (1) by 3;

30i₁ + 12i₂ = 60

Solving for i₁

30i₁ + 12i₂ = 60

8i₁ - 12i₂ = 12

38i₁ = 72

The above equation can be also simplified by Elimination or Cramer’s Rule.

i₁ = 72/38 = 1.895 Ampere = Current in 10 Ohms resistor

Substituting this value in (1), we get:

10(1.895) + 4i₂ = 20

4i₂ = 20 - 18.95

i₂ = 0.263 Amperes = Current in 4 Ohms Resistors.

Now,

i₁ - i₂ = 1.895 - 0.263 = 1.632 Amperes

Ans. In this solution, the direction of the current is the same as the direction of clockwise rotation.

(-20 - 5I - 5I - 12 - 10I) = 0

(-32 - 20I) = 0

(-32) = 20I

I = -32 / 20

I = -1.6 A

Because the electric current is negative, the direction of the electric current is actually opposite to the clockwise direction. The direction of electric current is not the same as estimation.

Ans. Resistor 1 (R₁) and resistor 2 (R₂) are connected in parallel. The equivalent resistor :

1/R₁₂ = 1/R₁ + 1/R₂ = 1/12 + 1/12 = 2/12

R₁₂ = 12/2 = 6 Ω

In this solution, the direction of current is the same as the direction of clockwise rotation.

(- I R₁₂ - E₁ - I R₃ - I R₄ + E₂ )= 0

(- 6 I - 6 - 3I - 6I + 12) = 0

(- 6I - 3I - 6I) = 6 -12

(- 15I ) = - 6

I = -6/-15

I = 2/5 A