Question 530: Current Electricity

Question

A potentiometer wire has length of $5 \mathrm{~m}$ and resistance of $16 \Omega$. The driving cell has an e.m.f. of $5 \mathrm{~V}$ and an internal resistance of $4 \Omega$. When the two cells of e.m.f.s $1.3 \mathrm{~V}$ and $1.1 \mathrm{~V}$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively

Accepted Answer

$\begin{aligned}
& K=\frac{E R}{(R+r) L} \
& \mathrm{E}=5 \mathrm{~V}, \mathrm{r}=4 \Omega, \mathrm{L}=5 \mathrm{~m}, \mathrm{R}=16 \Omega \
& 	herefore \quad \mathrm{K}=\frac{5 	imes 16}{(16+4) 	imes 5} \
& 	herefore \quad \mathrm{K}=0.8 \mathrm{~V} / \mathrm{m} \
\end{aligned}$
When '$E_1$' and '$E_2$' are connected so as to assist each other
$\begin{array}{ll} 
& \mathrm{E}_1+\mathrm{E}_2=\mathrm{K}_1 \
& 1.3+1.1=0.8 	imes l_1 \
	herefore \quad & l_1=3 \mathrm{~m}
\end{array}$
When '$E_1$' and '$E_2$' are connected so as to oppose each other,
$\begin{aligned}
& \mathrm{E}_1-\mathrm{E}_2=\mathrm{K} l_2 \
& 1.3-1.1=0.8 	imes l_2 \
	herefore \quad & l_2=0.25 \mathrm{~m}
\end{aligned}$
As, value for balancing lengths are different in all the options. It is sufficient to calculate balancing length in any one case (Assisting/ opposing) to reach the final correct answer.

Answer

$3 \mathrm{~m}, 0.25 \mathrm{~m}$

Answer

$0.25 \mathrm{~m}, 3 \mathrm{~m}$

Answer

$2.5 \mathrm{~m}, 0.3 \mathrm{~m}$

Answer

$0.3 \mathrm{~m}, 2.5 \mathrm{~m}$

Current Electricity

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