Kirchhoff's Laws & Electric Devices

1 Mark Questions

1.A heating element is marked 210 V, 630 What is the value of the current drawn by the element when connected to a 210 V DC Source? [Delhi 2013] Ans.

Given that <math>P = 630 \text{ W}</math> and <math>V = 210 \text{ V}</math>. In DC source <math>P = VI</math>. Therefore, <math>I = \frac{P}{V} = \frac{630}{210} = 3</math> A. (1)

2.In an experiment on meter bridge, if the balancing length AC is X, what would be its value, when the radius of the meter bridge wire AB is doubled? Justify your answer.[All India 2011 C]

Ans.

The balancing length continue to be X even on doubling the radius of meter bridge wire as it does not affect the ratio of length of two parts of meter bridge wire. (1)

NOTE : Resistance of wire <math>=</math> <math>\left(\frac{\rho}{A}\right)l</math><br>For uniform wire, <math>\left(\frac{\rho}{A}\right)</math> is constant even on doubling the radius of meter bridge wire. ∴ Resistance of wire <math>\infty</math> l.

3.In a meter bridge, two unknown resistances R and S when connected in the two gaps, give a null point at 40 cm from one end. What is the ratio of R and S?[Delhi 2010] Ans.

: Null point is obtained at 40 cm from one end

<math>l = 40 \text{ cm},</math> <math>100 - l = 60</math> cm.

: For meter bridge ratio of unknown resistances

<math display="block">\frac{R}{S} = \frac{l}{(100 - l)} = \frac{40}{60} = \frac{2}{3}</math> <math>\Rightarrow R:S=2:3</math> (1) 4.Use Kirchhoff's rules to determine the value of the current I_x flowing in the circuit shown in the figure.

<math>30 \Omega</math> <math>I_1</math> ₩₩

<math>I_1</math> <math>20 \Omega</math> 20 V ww

<math>I_2</math> ₩ 80 V <math>20 \Omega</math> <math>I_2</math> [Delhi 2013C]

Ans.

According to the question,

<math>30 \Omega</math> ~~~ В

<math>I_1</math> <math>20 \Omega</math> 20 V F ~~~ C <math>I_3</math> <math>I_2</math>

E n 80 V 20 Ω <math>I_2</math>

Applying Kirchhoff's junction rule at F node

<math>I_3 = I_1 + I_2</math>

...(i)

Applying Kirchhoff's second rule in loop ABCF,

<math>-30I_1 + 20 - 20I_3 = 0</math> <math>3I_1 + 2I_3 = 2</math>

...(ii)

In loop ABDE, (1) <math>-30I_1 + 20I_2 - 80 = 0</math> <math>-3I_1 + 2I_2 = 8</math>

...(iii)

From Eq. (i) put the value of <math>I_3</math> in Eq. (ii),

<math>3I_1 + 2I_1 + 2I_2 = 2</math> <math>5I_1 + 2I_2 = 2</math>

...(iv)

...(v)

subtract

<math display="block">-3I_1 + 2I_2 = 8</math>

<math display="block">8I_1 = -6, I_1 = -\frac{3}{4} A</math>

Put <math>I_1</math> in Eq. (iv)

<math>-5 \times \frac{3}{4} + 2I_2 = 2</math>, <math>2I_2 = \frac{23}{4}</math>, <math>I_2 = \frac{23}{8}</math> A

From Eq. (i)

<math>I_3 = -\frac{3}{4} + \frac{23}{8} = \frac{-6 + 23}{8}</math> or <math>I_3 = \frac{17}{8}</math> A. (1)