Question

When 2 resistances P and Q are kept in the left and right gaps of a meter bridge, the null point is obtained at 60 cm. If P is shunted by a resistance equal to half of its value, the shift in null point is

Answer 1

Answer:

Null point is shift by 26.7 cm towards left

Explanation:

As we know by condition of Null point we have

$\frac{R_p}{R_q} = \frac{60}{40}$

now we have shunted the P resistance with another resistance of half value

so we have

$R = \frac{R_p \times (0.5R_p)}{R_p + 0.5 R_p}$

$R = \frac{R_p}{3}$

now again we have

$\frac{R_p/3}{R_q} = \frac{x}{100 - x}$

now we have

$\frac{1}{3}(\frac{3}{2}) = \frac{x}{100 - x}$

so we have

$2x = 100 - x$

$x = 33.3 cm$

so shift of null point is given as

$\Delta x = 60 - 33.3$

$\Delta x = 26.7 cm$

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Topic : Potentiometer

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