A potentiometer wire has a length of 5 , m an | vedclass

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(A) The potential gradient $K$ of the potentiometer wire is given by:$K = \frac{V_{wire}}{L} = \frac{E_{driving} 	imes R}{(R + r) L}$Given $E_{drivin

Vedclass · Accepted Answer

$3 \, m, 0.25 \, m$

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(A) The potential gradient $K$ of the potentiometer wire is given by:$K = \frac{V_{wire}}{L} = \frac{E_{driving} 	imes R}{(R + r) L}$Given $E_{driving} = 5 \, V$, $r = 4 \, \Omega$, $L = 5 \, m$, and $R = 16 \, \Omega$:$K = \frac{5 	imes 16}{(16 + 4) 	imes 5} = \frac{80}{100} = 0.8 \, V/m$Case $1$: Cells assist each other $(E_{net} = E_1 + E_2)$$E_1 + E_2 = K l_1$$1.3 + 1.1 = 0.8 	imes l_1$$2.4 = 0.8 	imes l_1 \implies l_1 = 3 \, m$Case $2$: Cells oppose each other $(E_{net} = E_1 - E_2)$$E_1 - E_2 = K l_2$$1.3 - 1.1 = 0.8 	imes l_2$$0.2 = 0.8 	imes l_2 \implies l_2 = 0.25 \, m$Thus, the balancing lengths are $3 \, m$ and $0.25 \, m$.

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