Find the new position of the balance point when 5 - resistor is changed by 4 - resistor

The correct option is D 67.2 cm r = l_1 l_2l_2 R = 84 7070× 5 = 1 Ω 1 = l_1 l_3l_3× 4, #160; l_1 = 84 cm and get l_3 = 67.2 cm ...

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Standard XII

Physics

Potentiometer

Find the new ...

Question

Find the new position of the balance point when $5 ω$ resistor is changed by 4 $ω$ resistor

26.5 cm

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52 cm

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67.2 cm

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83.3 cm

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Solution

The correct option is D 67.2 cm
$r = \frac{l_{1} - l_{2}}{l_{2}} R = \frac{84 - 70}{70} \times 5 = 1 Ω$

$1 = \frac{l_{1} - l_{3}}{l_{3}} \times 4,$

$l_{1} = 84$ cm and get $l_{3} = 67.2 c m$

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Find the new position of the balance point when 5ω resistor is changed by 4 ω resistor

The correct option is D 67.2 cmr=l1−l2l2R=84−7070×5=1Ω1=l1−l3l3×4, l1=84 cm and get l3=67.2cm

Find the new position of the balance point when $5 ω$ resistor is changed by 4 $ω$ resistor

The correct option is D 67.2 cm
$r = \frac{l_{1} - l_{2}}{l_{2}} R = \frac{84 - 70}{70} \times 5 = 1 Ω$

$1 = \frac{l_{1} - l_{3}}{l_{3}} \times 4,$

$l_{1} = 84$ cm and get $l_{3} = 67.2 c m$