CBSE Class 12 · 2019 · Current ElectricityShort Answer

Using Kirchhoff's rules, calculate the current through the

40 Ω

and

20 Ω

resistors in the given circuit. \\ \textbf{OR} \\ What is end error in a metre bridge and how is it overcome? The two arms have

R = 5 Ω

and

S

. When

S

is shunted with an equal resistance, the new balance length is

1.5 ℓ_{1}

(with

ℓ_{1}

the initial length). Calculate

S

Answer: Current through the $20 Ω$ resistor $= 4 A$ ; through the $40 Ω$ resistor $= 0$ . [OR] $S = 10 Ω$ .

Solution

Applying the loop rule to the two meshes:

- 10 (I_{1} + I_{2}) + 40 - 40 I_{1} = 0 \Rightarrow 4 - 5 I_{1} = I_{2}

and

40 I_{1} + 80 - 20 I_{2} = 0 \Rightarrow 2 I_{1} + 4 = I_{2}

. Solving,

I_{1} = 0

I_{2} = 4 A

— so

4 A

flows through the

20 Ω

resistor and none through the

40 Ω

resistor. \textbf{OR:} end error is the small error from the stray resistances and the non-coincidence of the wire ends with the scale zeros; it is removed by interchanging R and S (and averaging) or by end corrections. With

\frac{R}{S} = \frac{ℓ _{1}}{100 - ℓ _{1}}

and

\frac{2 R}{S} = \frac{1.5 ℓ _{1}}{100 - 1.5 ℓ _{1}}

, solving gives

S = 2 R = 10 Ω.

CBSE Class 12 Board · 2019 · Subjective

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CBSE Class 12 · 2019 · Current ElectricityShort Answer

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