In a potentiometer experiment, balancing length is found to be $120$ cm for a cell $E_{1}$ of emf $2 V$ . What will be the balancing length for another cell $E_{2}$ of emf $1.5$ V? (No other changes are made in the experiment.)

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Solution

Given : $E_{1} = 2 V$ $l_{1} = 120 c m$ $E_{2} = 1.5 V$
Let the balancing length for $E_{2}$ be $l_{2}$ .
Using $\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}}$
$∴$ $\frac{1.5}{2} = \frac{l_{2}}{120}$
$⟹ l_{2} = 90 c m$

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In a potentiometer experiment, balancing length is found to be 120cm for a cell E1 of emf 2V. What will be the balancing length for another cell E2 of emf 1.5V? (No other changes are made in the experiment.)

Given : E1=2 V l1=120 cm E2=1.5 VLet the balancing length for E2 be l2.Using E2E1=l2l1∴ 1.52=l2120⟹ l2=90 cm

In a potentiometer experiment, balancing length is found to be $120$ cm for a cell $E_{1}$ of emf $2 V$ . What will be the balancing length for another cell $E_{2}$ of emf $1.5$ V? (No other changes are made in the experiment.)

Given : $E_{1} = 2 V$ $l_{1} = 120 c m$ $E_{2} = 1.5 V$
Let the balancing length for $E_{2}$ be $l_{2}$ .
Using $\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}}$
$∴$ $\frac{1.5}{2} = \frac{l_{2}}{120}$
$⟹ l_{2} = 90 c m$