Potentiometer Questions in English

(C) $1$. First, calculate the total resistance of the circuit: $R_{total} = R_{wire} + R_{series} + r = 5 \Omega + 992 \Omega + 3 \Omega = 1000 \Omega$.
$2$. Calculate the current flowing through the potentiometer wire: $I = \frac{E}{R_{total}} = \frac{4 \text{ V}}{1000 \Omega} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
$3$. Calculate the potential drop across the entire potentiometer wire: $V_{wire} = I \times R_{wire} = 4 \times 10^{-3} \text{ A} \times 5 \Omega = 20 \times 10^{-3} \text{ V} = 20 \text{ mV}$.
$4$. The potential gradient $(k)$ along the wire is: $k = \frac{V_{wire}}{L} = \frac{20 \text{ mV}}{4 \text{ m}} = 5 \text{ mV/m}$.
$5$. The e.m.f. balanced by a length of $0.75 \text{ m}$ is: $E' = k \times l = 5 \text{ mV/m} \times 0.75 \text{ m} = 3.75 \text{ mV}$.

(C) In a potentiometer, the $E$.$M$.$F$. of a cell is directly proportional to the balancing length, i.e.,$E = k \cdot l$, where $k$ is the potential gradient.
$1$. When the cell $E_1$ is connected, the balancing length is $l_1$, so $E_1 = k \cdot l_1$.
$2$. When the cells $E_1$ and $E_2$ are connected in opposition, the effective $E$.$M$.$F$. is $(E_1 - E_2)$, and the balancing length is $l_2$, so $(E_1 - E_2) = k \cdot l_2$.
$3$. Dividing the two equations: $\frac{E_1}{E_1 - E_2} = \frac{k \cdot l_1}{k \cdot l_2} = \frac{l_1}{l_2}$.
$4$. Rearranging the equation: $E_1 \cdot l_2 = l_1 \cdot (E_1 - E_2) \implies E_1 \cdot l_2 = E_1 \cdot l_1 - E_2 \cdot l_1$.
$5$. Grouping terms: $E_2 \cdot l_1 = E_1 \cdot l_1 - E_1 \cdot l_2 = E_1(l_1 - l_2)$.
$6$. Therefore, $\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$.
Thus, the ratio $E_1 : E_2$ is $(l_1) : (l_1 - l_2)$.

(D) Let the e.m.f.s of the two cells be $E_1 = 1.5 \text{ V}$ and $E_2 = 1.2 \text{ V}$.
When the cells are connected to assist each other, the total e.m.f. is $E_{eq1} = E_1 + E_2 = 1.5 + 1.2 = 2.7 \text{ V}$.
The balancing length $l_1 = 270 \text{ cm}$.
In a potentiometer, $E \propto l$, so $E_{eq1} = k l_1$, where $k$ is the potential gradient.
$2.7 = k \times 270 \implies k = \frac{2.7}{270} = 0.01 \text{ V/cm}$.
When the cells are connected in opposition, the total e.m.f. is $E_{eq2} = E_1 - E_2 = 1.5 - 1.2 = 0.3 \text{ V}$.
Let the new balancing length be $l_2$.
Then $E_{eq2} = k l_2$.
$0.3 = 0.01 \times l_2$.
$l_2 = \frac{0.3}{0.01} = 30 \text{ cm}$.

(C) In a potentiometer, the balancing length $l$ is directly proportional to the e.m.f. $E$ of the cell, i.e.,$E = kl$, where $k$ is the potential gradient of the potentiometer wire.
When the potentiometer is connected between points $A$ and $B$, the e.m.f. measured is $E_1$. Given the balancing length $l_1 = 412 \text{ cm}$, we have:
$E_1 = k \times 412$ --- $(1)$
When the potentiometer is connected between points $A$ and $C$, the total e.m.f. measured is the sum of the two cells connected in series, which is $E_1 + E_2$. Given the balancing length $l_2 = 103 \text{ cm}$, we have:
$E_1 + E_2 = k \times 103$ --- $(2)$
Wait, looking at the circuit diagram, the cells are connected in opposition (positive terminal of $E_1$ faces positive terminal of $E_2$). Thus, the effective e.m.f. between $A$ and $C$ is $E_1 - E_2$.
$E_1 - E_2 = k \times 103$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{E_1}{E_1 - E_2} = \frac{412}{103}$
$\frac{E_1}{E_1 - E_2} = 4$
$E_1 = 4(E_1 - E_2)$
$E_1 = 4E_1 - 4E_2$
$3E_1 = 4E_2$
$\frac{E_1}{E_2} = \frac{4}{3}$
Therefore, the ratio $E_1 : E_2$ is $4 : 3$.

(A) $1$. First, calculate the current $I$ flowing through the potentiometer wire $AB$ in the primary circuit:
$I = \frac{E}{R_{total}} = \frac{5 \text{ V}}{45 \text{ } \Omega + 5 \text{ } \Omega} = \frac{5}{50} \text{ A} = 0.1 \text{ A}$.
$2$. Calculate the potential drop across the potentiometer wire $AB$:
$V_{AB} = I \times R_{AB} = 0.1 \text{ A} \times 5 \text{ } \Omega = 0.5 \text{ V}$.
$3$. The potential gradient $k$ along the wire is given by:
$k = \frac{V_{AB}}{L} = \frac{0.5 \text{ V}}{10 \text{ m}} = 0.05 \text{ V/m}$.
$4$. For the balancing length $AP = l$, the potential drop across $AP$ must equal the e.m.f. of the secondary cell $(E' = 0.4 \text{ V})$:
$V_{AP} = k \times l = 0.4 \text{ V}$.
$0.05 \text{ V/m} \times l = 0.4 \text{ V}$.
$l = \frac{0.4}{0.05} \text{ m} = 8 \text{ m}$.
Therefore, the balancing length $AP$ is $8 \text{ m}$.