In the potentiometer circuit shown in figure, balance is obtained. The unknown $E_{x}$ is then

200 mV

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2.8 V

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3.0 V

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400 mV

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Solution

The correct option is A 200 mV

for the case of no-loading $(I_{G} = 0)$

$∴ V_{A B} = E_{x}$

$V_{A B} = \frac{200}{200 + 2800 + 200} \times 3.2$

= 0.2 V
= 200 mV

$∴ E_{x} = 200 m V$

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In the potentiometer circuit shown in figure, balance is obtained. The unknown Ex is then

The correct option is A 200 mV for the case of no-loading (IG=0) ∴ VAB=Ex VAB=200200+2800+200×3.2 = 0.2 V = 200 mV ∴ Ex=200 mV

In the potentiometer circuit shown in figure, balance is obtained. The unknown $E_{x}$ is then

The correct option is A 200 mV

for the case of no-loading $(I_{G} = 0)$

$∴ V_{A B} = E_{x}$

$V_{A B} = \frac{200}{200 + 2800 + 200} \times 3.2$

= 0.2 V
= 200 mV

$∴ E_{x} = 200 m V$