Mix Examples-Current Electricity Questions in English

(B) The relationship between the temperature of inversion $(T_i)$,neutral temperature $(T_n)$,and the temperature of the cold junction $(T_0)$ is given by the formula:
$T_n = \frac{T_i + T_0}{2}$
Rearranging the formula to solve for $T_i$:
$T_i = 2T_n - T_0$
Given values are $T_0 = 10^{\circ} C$ and $T_n = 270^{\circ} C$.
Substituting these values into the equation:
$T_i = 2 \times 270^{\circ} C - 10^{\circ} C$
$T_i = 540^{\circ} C - 10^{\circ} C$
$T_i = 530^{\circ} C$
Therefore,the temperature of inversion is $530^{\circ} C$.

(C) The current $i$ flowing through the circuit is given by the net emf divided by the total resistance of the circuit.
Since the cells are connected in opposition,the net emf is $\varepsilon_{net} = 2 \text{ V} - 1.5 \text{ V} = 0.5 \text{ V}$.
The total resistance of the circuit is $R_{total} = 5 \text{ } \Omega + 5 \text{ } \Omega + 10 \text{ } \Omega = 20 \text{ } \Omega$.
Therefore,the current $i = \frac{0.5 \text{ V}}{20 \text{ } \Omega} = 0.025 \text{ A}$.
For cell $A$,which is discharging,the terminal potential difference is $V_{A} = \varepsilon_{A} - i r_{A} = 2 \text{ V} - (0.025 \text{ A} \times 5 \text{ } \Omega) = 2 - 0.125 = 1.875 \text{ V}$.
For cell $B$,which is charging,the terminal potential difference is $V_{B} = \varepsilon_{B} + i r_{B} = 1.5 \text{ V} + (0.025 \text{ A} \times 5 \text{ } \Omega) = 1.5 + 0.125 = 1.625 \text{ V}$.

(B) The heat generated in a resistance $R$ connected to a cell of emf $E$ and internal resistance $r$ is given by $H = I^2 R t = \left( \frac{E}{R+r} \right)^2 R t$.
Given $H_1 = H$ for $R_1$ and $H_2 = 4H$ for $R_2$ in the same time $t$,we have:
$H = \frac{E^2 R_1}{(R_1+r)^2} t$ and $4H = \frac{E^2 R_2}{(R_2+r)^2} t$.
Dividing the two equations: $\frac{4H}{H} = \frac{R_2 (R_1+r)^2}{R_1 (R_2+r)^2} \Rightarrow 4 = \frac{R_2 (R_1+r)^2}{R_1 (R_2+r)^2}$.
Taking the square root on both sides: $2 = \frac{\sqrt{R_2} (R_1+r)}{\sqrt{R_1} (R_2+r)}$.
$2 \sqrt{R_1} (R_2+r) = \sqrt{R_2} (R_1+r) \Rightarrow 2 \sqrt{R_1} R_2 + 2 \sqrt{R_1} r = \sqrt{R_2} R_1 + \sqrt{R_2} r$.
Rearranging for $r$: $r (2 \sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_2} R_1 - 2 \sqrt{R_1} R_2$.
$r = \frac{\sqrt{R_1 R_2} (\sqrt{R_1} - 2 \sqrt{R_2})}{2 \sqrt{R_1} - \sqrt{R_2}}$.
Multiplying numerator and denominator by $-1$: $r = \sqrt{R_1 R_2} \frac{2 \sqrt{R_2} - \sqrt{R_1}}{\sqrt{R_2} - 2 \sqrt{R_1}}$.

(B) $1$. First,calculate the total resistance of the circuit. The resistors $1 \text{ k}\Omega$ and $2 \text{ k}\Omega$ are in series with the $3 \text{ k}\Omega$ resistor. Total resistance $R_{eq} = 1 \text{ k}\Omega + 2 \text{ k}\Omega + 3 \text{ k}\Omega = 6 \text{ k}\Omega = 6000 \ \Omega$.
$2$. The current $I$ flowing through the circuit is $I = \frac{E}{R_{eq}} = \frac{3 \text{ V}}{6000 \ \Omega} = 0.5 \times 10^{-3} \text{ A} = 0.5 \text{ mA}$.
$3$. The potential at point $B$ relative to $A$ is $V_{AB} = I \times R_{1k} = 0.5 \text{ mA} \times 1 \text{ k}\Omega = 0.5 \text{ V}$. Thus,$V_A - V_B = 0.5 \text{ V}$.
$4$. The potential at point $C$ relative to $A$ is determined by the voltage divider rule for the capacitors. Since the capacitors are in series and no $DC$ current flows through them,the potential $V_C$ is determined by the ratio of their capacitances. $V_A - V_C = V_{AD} \times \frac{C_2}{C_1 + C_2}$,where $V_{AD} = I \times (1 \text{ k}\Omega + 2 \text{ k}\Omega) = 0.5 \text{ mA} \times 3 \text{ k}\Omega = 1.5 \text{ V}$.
$5$. $V_A - V_C = 1.5 \text{ V} \times \frac{2 \mu\text{F}}{1 \mu\text{F} + 2 \mu\text{F}} = 1.5 \times \frac{2}{3} = 1.0 \text{ V}$.
$6$. Now,$V_{BC} = V_B - V_C = (V_A - V_C) - (V_A - V_B) = 1.0 \text{ V} - 0.5 \text{ V} = 0.5 \text{ V}$.