Question

A 'Wheatstone Bridge' circuit has been set up as shown. The resistor $R_4$ is an ideal carbon resistance (tolerance$=0\%$) having bands of colours black, yellow and brown marked on it. The galvanometer, in this circuit, would show a 'null point' when another ideal carbon resistor $X$ is connected across $R_4,$ having bands of colours

• A. Black, brown, black is put in parallel with $R_4$
• B. Black, brown, brown is put in parallel with $R_4$
• C. Brown black, brown is put in series with $R_4$
• D. Black, brown, black is put in series with $R_4$

Answer 1

The correct option is B Black, brown, brown is put in parallel with $R_4$

Color digit multiplier

Black $0$ $10$

Brown $1$ ${10}^1$

Yellow $4$ ${10}^4$

Ref. image I

we have resistance $R_4$ as

Ref. image II

$R = digit X multiplier

$R_4=04\times {10}^1$

$\Rightarrow R_4=40\Omega$

Ref image III

Let us take the net resistance of co after connecting resistor X e 'R'

Since,

It is when stone bridge

$\Rightarrow \frac{BC}{CD}=\frac{AB}{AD}$

$\Rightarrow \frac{R_3}{R_4}=\frac{R_1}{R_2}$

$\Rightarrow \frac{24}{R}=\frac{48}{16}$

$\Rightarrow R=\frac{24\times 16}{46}$

$\Rightarrow R=2\Omega <40\Omega$

Since $R<R_4$

So, the resistor X will be in parallel to $R_4$

$\Rightarrow \frac{1}{R_4}+\frac{1}{X}=\frac{1}{R}$

$\Rightarrow \frac{1}{40}+\frac{1}{x}=\frac{1}{6}$

$\Rightarrow \frac{1}{X}=\frac{1}{8}-\frac{1}{40}=\frac{4}{40}=\frac{1}{10}$

So, option B i.e. Black, brown, brown in parallel to $R_4$ is correct.

Ref image IV

$R=01\times {10}^1=10\Omega$