Current Electricity

The given circuit is a symmetric unbalanced bridge.

To find the equivalent resistance ( $R_{A B}$ ), we apply a potential difference $V$ across terminals $A$ and $B$ and determine the total current I entering the circuit.

Let the potential at node A be V (input terminal) and the potential at node B be 0.

Using Kirchhoff's Current Law (KCL) the sum of currents leaving each junction is zero.

At node $\mathrm{V}_1$ :

$ \frac{\left(V_1-V\right)}{6}+\frac{\left(V_1-0\right)}{3}+\frac{\left(V_1-V_2\right)}{3}=0 $

$\Rightarrow $ $\left(V_1-V\right)+2 V_1+2\left(V_1-V_2\right)=0$

$\Rightarrow $ $ 5 V_1-2 V_2=V \ldots \text { (i) } $

At node $\mathrm{V}_2$ :

$\frac{V_2-V}{3}+\frac{V_2-0}{6}+\frac{V_2-V_1}{3}=0$

$\Rightarrow $ $2\left(V_2-V\right)+V_2+2\left(V_2-V_1\right)=0$

$\Rightarrow $ $ -2 \mathrm{~V}_1+5 \mathrm{~V}_2=2 \mathrm{~V} \ldots \text { (ii) } $

Solving both the equations, (i) $\times 5+$ (ii) $\times 2$

$ \left(5 V_1-2 V_2\right) \times 5+\left(-2 V_1+5 V_2\right) \times 2=V \times 5+2 V \times 2 $

$\Rightarrow $ $\left(25 V_1-10 V_2\right)+\left(-4 V_1+10 V_2\right)=5 V+4 V$

$\Rightarrow $ $21 V_1=9 V \Rightarrow V_1=\frac{9}{21} V=\frac{3}{7} V$

Substituting $V_1$ into Equation (i) to find $V_2$ :

$ \begin{gathered} 5\left(\frac{3}{7} \mathrm{~V}\right)-2 \mathrm{~V}_2=\mathrm{V} \\ 2 \mathrm{~V}_2=\frac{15}{7} \mathrm{~V}-\mathrm{V}=\frac{8}{7} \mathrm{~V} \Rightarrow \mathrm{~V}_2=\frac{4}{7} \mathrm{~V} \end{gathered} $

The total current I entering from terminal A is the sum of currents in the two left branches :

$ I=\frac{V-V_1}{6}+\frac{V-V_2}{3} $

$\Rightarrow $ $I=\frac{V-\frac{3}{7} V}{6}+\frac{V-\frac{4}{7} V}{3}$

$\Rightarrow $ $I=\frac{\frac{4 V}{7}}{6}+\frac{\frac{3 V}{7}}{3}=\frac{2 V}{21}+\frac{V}{7}$

$ I=\frac{2 V+3 V}{21}=\frac{5 V}{21} $

The equivalent resistance $R_{A B}$ is :

$ R_{A B}=\frac{V}{I}=\frac{V}{5 V / 21}=\frac{21}{5} \Omega $

Given that $\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{x}}{5} \Omega :$

$ \frac{x}{5}=\frac{21}{5} \Rightarrow x=21 $

Therefore, the value of x is 21.

A meter bridge works on the principle of a balanced Wheatstone bridge. The ratio of the resistances in the gaps is equal to the ratio of the lengths of the wire segments.

Case 1 :

Null point from left is l cm . So, the remaining length from right (100 - l) cm

At balanced condition

$ \frac{2}{3}=\frac{l}{100-l} \Rightarrow 2(100-l)=3l $

$\Rightarrow 200-2l=3l$

$\Rightarrow 5 \mathrm{l}=200 \Rightarrow \mathrm{l}=40 \mathrm{~cm}$

Case 2 :

When an unknown resistance x is connected in parallel with the $3 \Omega$ resistor in the right gap. This changes the equivalent resistance of the right gap.

New right gap resistance $S^{\prime}=\frac{3 x}{3+x}$

New null point $\mathrm{l}^{\prime}=\mathrm{l}+10=40+10=50 \mathrm{~cm}$

So, the remaining length from right $=100-50=50 \mathrm{~cm}$

At balanced condition for the second case:

$ \frac{\mathrm{R}}{\mathrm{~S}^{\prime}}=\frac{\mathrm{l}^{\prime}}{100-\mathrm{l}^{\prime}} $

$\Rightarrow $ $\frac{2}{\left(\frac{3 x}{3+x}\right)}=\frac{50}{50} \Rightarrow \frac{6+2 x}{3 x}=1$

$ \Rightarrow 3 x=6+2 x \Rightarrow x=6 \Omega $

The value of unknown resistance x is $6 \Omega$.

Hence, the correct answer is 6.

The mobility $(\mu)$ of free electrons in a conductor is defined as the magnitude of their drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)$ per unit electric field (E).

$ \mu=\frac{v_d}{E} $

The electric field ( E ) established across a conductor of length L connected to a potential difference V is given by:

$ \mathrm{E}=\frac{\mathrm{V}}{\mathrm{~L}} $

The electric current (I) flowing through a conductor with cross-sectional area A and free electron number density n is:

$ \mathrm{I}=\mathrm{neA} \mathrm{v}_{\mathrm{d}} $

$ \Rightarrow \mathrm{v}_{\mathrm{d}}=\frac{1}{\mathrm{ne} \mathrm{~A}} $

Using mobility formula,

$\mu=\frac{\left(\frac{\mathrm{I}}{\mathrm{ne} \mathrm{A}}\right)}{\left(\frac{\mathrm{V}}{\mathrm{L}}\right)}$

$\Rightarrow $ $ \mu=\frac{\mathrm{IL}}{\mathrm{ne} \mathrm{AV}} $

We have given,

• Length, $\mathrm{L}=2 \mathrm{~m}$

• Current, $\mathrm{I}=1.6 \mathrm{~A}$

• Voltage, $\mathrm{V}=2 \mathrm{~V}$

• Electron concentration, $\mathrm{n}=5 \times 10^{28} \mathrm{~m}^{-3}$

• Electron charge, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$

• Area of cross-section, $\mathrm{A}=0.2 \mathrm{~mm}^2=0.2 \times 10^{-6} \mathrm{~m}^2=2 \times 10^{-7} \mathrm{~m}^2$

Substituting the values,

$ \mu=\frac{1.6 \times 2}{\left(5 \times 10^{28}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(2 \times 10^{-7}\right) \times 2} $

$\Rightarrow $ $\mu=\frac{3.2}{32 \times 10^2}$

$\Rightarrow $ $\mu=0.001 \mathrm{~m}^2 / \mathrm{V} \cdot \mathrm{s}$

$\Rightarrow $ $\mu=1 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V} \cdot \mathrm{s}$

$\Rightarrow 1 \times 10^{-3}=\alpha \times 10^{-3} \Rightarrow \alpha=1$

A bridge is balanced if the ratio of the resistances in the two arms is equal :

$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_3}{\mathrm{R}_4} $

So, from the given circuit, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{2}$ and $\frac{\mathrm{R}_3}{\mathrm{R}_4}=\frac{2}{4}=\frac{1}{2} \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_3}{\mathrm{R}_4}$

Since the ratio is equal $\left(\frac{1}{2}=\frac{1}{2}\right)$, the bridge is balanced.

Because the bridge is balanced, no current flows through the central $1 \Omega$ resistor. The circuit now simplifies to two parallel branches :

Here, $\mathrm{R}_1$ and $\mathrm{R}_3$ are in series, $\mathrm{R}_{13}=1+2=3 \Omega$

Also, $\mathrm{R}_2$ and $\mathrm{R}_4$ are in series, $\mathrm{R}_{24}=2+4=6 \Omega$

Now, $\mathrm{R}_{13}$ and $\mathrm{R}_{24}$ are in parallel.

So, the equivalent resistance between A and $\mathrm{B}\left(\mathrm{R}_{\mathrm{AB}}\right)$ is :

$ \frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{1}{2} $

$\Rightarrow $ $ \mathrm{R}_{\mathrm{AB}}=2 \Omega $

The battery has an emf of 9 V and an internal resistance (r) of $1 \Omega$. The total resistance ( $\mathrm{R}_{\text {total }}$ ) is:

$ \mathrm{R}_{\text {total }}=\mathrm{R}_{\mathrm{AB}}+\mathrm{r}=2+1=3 \Omega $

Using Ohm's Law for the whole circuit:

$ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {total }}}=\frac{9}{3}=3 \text { Amperes } $

So, the power dissipated between points A and B is,

$\mathrm{P}_{\mathrm{AB}}=\mathrm{I}^2 \cdot \mathrm{R}_{\mathrm{AB}} $

$\Rightarrow $ $ \mathrm{P}_{\mathrm{AB}}=(3)^2 \times 2=9 \times 2=18 \text { Watts } $

So, the total heat $(\mathrm{H})$ generated in $\mathrm{t}=1$ minute $=60$ seconds is :

$ \mathrm{H}=\mathrm{P}_{\mathrm{AB}} \times \mathrm{t} $

$\Rightarrow $ $ \mathrm{H}=18 \times 60=1080 \mathrm{~J} $

Therefore, the heat generated in one minute is 1080 J between points A and B .

Hence, the correct answer is 1080.

Series Connection:

Total emf = 1 V + 2 V = 3 V.

Total resistance = 2 Ω + 1 Ω + 6 Ω = 9 Ω.

Current $I_1 = \frac{\text{emf}}{\text{resistance}} = \frac{3}{9} = \tfrac{1}{3}\text{ A}.$

Parallel Connection:

Equivalent emf: $\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\tfrac{1}{2} + \tfrac{2}{1}}{\tfrac{1}{2} + 1} = \frac{5}{3}\text{ V}.$

Equivalent internal resistance of the cells: $r_{\text{int,eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\,\Omega.$

Total resistance = external 6 Ω + internal $\tfrac{2}{3}$ Ω = $\tfrac{20}{3}$ Ω.

Current $I_2 = \frac{\varepsilon_{\text{eq}}}{\text{total resistance}} = \frac{5/3}{20/3} = \tfrac{1}{4}\text{ A}.$

Finally, $\frac{I_1}{I_2} = \frac{\tfrac{1}{3}}{\tfrac{1}{4}} = \frac{4}{3},$ so $x = 4\,$.

Step 1: Find the Total Resistance

First, add up the resistor in series ($R_1 = 150.4~\Omega$) with the combined resistance of the ammeter and its shunt ($R_2$).

The ammeter ($240~\Omega$) is in parallel with a shunt of ($10~\Omega$). The combined resistance of resistors in parallel is found by:

$ R_2 = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6~\Omega $

The total resistance ($R_{eq}$) is:

$ R_{eq} = 150.4 + 9.6 = 160~\Omega $

Step 2: Find the Current in the Circuit

Suppose the total current in the circuit is $I$. But to find the ammeter reading, we care about the current through the ammeter coil itself.

The current through the ammeter, $I_1$, is given by how the current splits between the ammeter and the shunt. The fraction going through the ammeter is:

$ I_1 = \frac{R_2}{240} \times I $

But $R_2 = 9.6~\Omega$ (the parallel combination), and $240~\Omega$ is the ammeter's resistance:

$ I_1 = \frac{9.6}{240} \times I $

Step 3: Substitute the Circuit Values

The supply voltage is $20~V$. The total resistance is $160~\Omega$. So the total current is:

$ I = \frac{20}{160} = 0.125~A $

So: $ I_1 = 0.125 \times \frac{9.6}{240} $

Calculate this: $ I_1 = 0.125 \times 0.04 = 0.005~A = 5~\text{mA} $

Final Answer:

The ammeter reads $5~\text{mA}$.

$\begin{aligned} & \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \Rightarrow \text { the bridge is balanced } \\ & \Rightarrow \frac{10}{\mathrm{R}}=\frac{20}{40} \\ & \mathrm{R}=20 \Omega \\ & \mathrm{I}=\frac{40}{20}=2 \mathrm{~A} \end{aligned}$

$ \text{Each side of the triangle has a resistance of } R_{\text{side}} = \frac{9\,\Omega}{3} = 3\,\Omega. $

When measuring the equivalent resistance between any two vertices, there are two paths:

A direct path along one side with resistance:

$ R_1 = 3\,\Omega. $

An indirect path passing through the other two sides in series:

$ R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega. $

These two paths are in parallel, so the equivalent resistance is calculated by:

$ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3\,\Omega} + \frac{1}{6\,\Omega} = \frac{2}{6\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{6\,\Omega}. $

Thus,

$ R_{\text{eq}} = \frac{6\,\Omega}{3} = 2\,\Omega. $

The equivalent resistance across any two vertices of the equilateral triangle is therefore:

$ 2\,\Omega. $

$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=2 \Omega \\ & \mathrm{I}=\frac{2}{2}=1 \mathrm{~A} \end{aligned}$

$\begin{aligned} & R_{\text {net }}=\frac{1}{\text { Slope }}=\frac{8}{4} \times 10^3=2 \times 10^3 \\ & \Rightarrow 2=\frac{10 R}{10+R} \\ & \Rightarrow 20+2 R=10 R \\ & \Rightarrow 8 R=20 \\ & \Rightarrow R=2.5 \mathrm{k} \Omega \\ & \quad=2500 ~\Omega \end{aligned}$

We can start solving this problem by first understanding that the resistance of a material changes with temperature, and this change can be quantified using the temperature coefficient of resistance $ \alpha $. The relationship between the resistance of a material at any temperature $ T $ and its resistance at a reference temperature $ T_0 $ is given by the formula:

$ R = R_0(1 + \alpha(T - T_0)) $

Where:

• $ R $ is the resistance at temperature $ T $ (in this case, $ 62 \Omega $).
• $ R_0 $ is the resistance at reference temperature $ T_0 $ (in this case, $ 50 \Omega $).
• $ \alpha $ is the temperature coefficient of resistance (in this case, $ 2.4 \times 10^{-4} \, ^\circ\mathrm{C}^{-1} $).
• $ T $ is the unknown temperature we need to find.
• $ T_0 $ is the reference temperature, given as $ 27^\circ \mathrm{C} $.

By substituting the given values into the formula, we get:

$ 62 = 50(1 + 2.4 \times 10^{-4}(T - 27)) $

First, divide both sides of the equation by 50:

$ \frac{62}{50} = 1 + 2.4 \times 10^{-4}(T - 27) $

Then solve for $ T $:

$ 1.24 = 1 + 2.4 \times 10^{-4}(T - 27) $

$ 0.24 = 2.4 \times 10^{-4}(T - 27) $

$ \frac{0.24}{2.4 \times 10^{-4}} = T - 27 $

$ 1000 = T - 27 $

$ T = 1027 ^\circ\mathrm{C} $

Thus, the temperature of the element when its resistance is $ 62 \Omega $ is $ 1027^\circ\mathrm{C} $.

At $C$

$\begin{aligned} & \frac{v_1}{2}+\frac{v_1-5}{2}+\frac{v_1+10-v_2}{1}=0 \\ & v_2=2 v_1+\frac{15}{2} \quad \text{.... (i)} \end{aligned}$

At $A$

$\begin{aligned} & \frac{v_2-5}{4}+\frac{v_2}{4}+\frac{v_2-10-v_1}{1}=0 \\ & 6 v_2=4 v_1+45 \quad \text{.... (ii)} \end{aligned}$

$\begin{aligned} & \Rightarrow v_1=0 \\ & \text { and } v_2=\frac{15}{2} \\ & \therefore \quad i=\frac{5}{2} \mathrm{~A} \\ & \Rightarrow n=25 \end{aligned}$

$\begin{gathered} R_H=\frac{100 \times 100}{1000}=10 \Omega \\ i_H=\sqrt{\frac{62.5}{10}}=2.5 \mathrm{~A} \\ V_H=25 \mathrm{~V} \\ \Rightarrow \quad i_b=\frac{75}{10}=7.5 \mathrm{~A} \\ \Rightarrow \quad i_R=5 \mathrm{~A} \\ R=\frac{25}{5}=5 \Omega \end{gathered}$

To determine the temperature $t$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:

$ R_t = R_0 (1 + \alpha t) $

where:

• $R_t$ is the resistance at temperature $t$
• $R_0$ is the resistance at the reference temperature (usually $0^{\circ} \mathrm{C}$)
• $\alpha$ is the temperature coefficient of resistance
• $t$ is the temperature

We are given the following resistances:

• Resistance at $0^{\circ} \mathrm{C}$: $R_0 = 10 \Omega$
• Resistance at $100^{\circ} \mathrm{C}$: $R_{100} = 10.2 \Omega$
• Resistance at $t^{\circ} \mathrm{C}$: $R_t = 10.95 \Omega$

First, we need to find the temperature coefficient of resistance $\alpha$. Using the resistance at $100^{\circ} \mathrm{C}$:

$ 10.2 = 10 (1 + \alpha \cdot 100) $

Solving for $\alpha$:

$ \frac{10.2}{10} = 1 + 100\alpha \implies 1.02 = 1 + 100\alpha \implies 100\alpha = 0.02 \implies \alpha = \frac{0.02}{100} = 0.0002 $

Now, we can find the temperature $t$ using the resistance at $t^{\circ} \mathrm{C}$:

$ 10.95 = 10 (1 + 0.0002 \cdot t) $

Solving for $t$:

$ \frac{10.95}{10} = 1 + 0.0002 t \implies 1.095 = 1 + 0.0002 t \implies 0.0002 t = 0.095 \implies t = \frac{0.095}{0.0002} = 475 $

The temperature $t$ in Celsius is $475^{\circ} \mathrm{C}$. To convert this to the Kelvin scale:

$ T_{K} = t_{C} + 273.15 = 475 + 273.15 = 748.15 \, \mathrm{K} $

So, the temperature $t$ in Kelvin scale is approximately $748.15 \, \mathrm{K}$.

$\begin{aligned} & i=\frac{24}{140 \cdot 4+r_A}=\frac{24}{140 \cdot 4+9 \cdot 6}=0.16 \mathrm{~A} \\ & i=160 \mathrm{~mA} \end{aligned}$

The resistance $R$ of a wire is given by the formula:

where:

• $\rho$ is the resistivity of the material,
• $l$ is the length of the wire,
• $A$ is the cross-sectional area of the wire.

If we have a cylindrical wire, the cross-sectional area can be expressed as $A = \pi r^2$, where $r$ is the radius of the cylinder. Therefore, the resistance of the original wire can be written as:

When the wire is stretched such that its radius becomes $r / 2$, its volume would remain constant, given that the volume of a cylinder is $V = A \cdot l = \pi r^2 \cdot l$. Assuming the volume before and after the stretching is the same, and since the area is now a quarter of the original (because when the radius is halved, the area, which is proportional to the square of the radius, is reduced to a quarter), the length must have increased to four times the original to preserve the volume. That is,

and the new area,

Therefore, the new resistance, $R'$, of the wire can be calculated using the original formula for resistance:

Hence, the new resistance of the wire is $16R$, which means $x = 16$.

Let's start by understanding the process of dividing the wire and recombining its parts to form the final configuration. Initially, we have a wire with a resistance of $20 \Omega$. This wire is divided into 10 equal parts, each part then has a resistance of:

$\frac{20 \Omega}{10} = 2 \Omega$

Since each part has the same length and presumably the same material and cross-sectional area, then each part will have the same resistance of $2 \Omega$.

When two parts are connected in parallel, the equivalent resistance, $R_{\text{parallel}}$, of this configuration can be calculated using the formula for two resistors in parallel:

$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$

Given that $R_1 = R_2 = 2 \Omega$ (since the parts are identical), we have:

$\frac{1}{R_{\text{parallel}}} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} = \frac{2}{2 \Omega}$

This simplifies to:

$\frac{1}{R_{\text{parallel}}} = \frac{2}{2 \Omega} = \frac{1}{\Omega}$

From which it follows that:

$R_{\text{parallel}} = 1 \Omega$

Now, since the original wire was divided into 10 equal parts, and pairs of these parts are connected in parallel, this results in $\frac{10}{2} = 5$ pairs. Each of these pairs has an equivalent resistance of $1 \Omega$.

Finally, these pairs are all connected in series. The total resistance of resistors in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of the final configuration, $R_{\text{series}}$, is:

$R_{\text{series}} = 5 \times R_{\text{parallel}} = 5 \times 1 \Omega = 5 \Omega$

So, the equivalent resistance of the final combination is $5 \Omega$.

$\frac{1}{3} \mathrm{~A} \longrightarrow \frac{1}{2} \mathrm{div}$

$\frac{1}{2} \mathrm{~A} \longrightarrow \frac{2}{3} \mathrm{div}$

$\text { Figure of merit }=\frac{\Delta i}{\Delta \theta} \quad \begin{aligned} & \frac{\frac{1}{2}--}{\frac{2}{3}-\frac{1}{2}} \\ &=0.5 \\ &=5 \times 10^{-1} \mathrm{~A} / \mathrm{div} \end{aligned}$

$\begin{aligned} & R=\rho \frac{I}{A}=\rho \frac{V}{A^2} \\ & \text { and } \pi r_1^2 I_1=\pi r_2^2 I_2 \\ & A_1 I_1=A_2 I_2 \\ & \text { So } \frac{R_1}{R_2}=\left(\frac{A_2}{A_1}\right)^2 \\ & \Rightarrow \frac{R}{2}=\left(\frac{r_2}{r_1}\right)^4 \\ & \Rightarrow R=32 \end{aligned}$

The circuit can be simplified as

$\begin{aligned} & R_{a c}=\frac{6 \times 2}{8}=\frac{3}{2} \Omega \\ & \begin{aligned} & i=1 \mathrm{Amp} . \\ & V_{e f}=\left(\frac{i}{2}\right)^2 \\ & \quad=1 \mathrm{~V} \end{aligned} \end{aligned}$

To find the amount of electric charge that flows through a section of the conductor, we have to integrate the current over the given time interval. The current $I(t)$ as a function of time $t$ is given by:

The electric charge $Q$ that flows through the conductor from time $t = 1$ s to $t = 2$ s is calculated by integrating the current $I(t)$ with respect to time over this interval:

Substituting the given limits ($t_1=1$ and $t_2=2$) and the expression for $I(t)$, we get:

Now we'll integrate the function with respect to $t$:

Simplifying the integrated function:

Substitute the upper and lower limits of the integration:

Therefore, the amount of electric charge that flows through the section of the conductor from $t=1$ s to $t=2$ s is 22 Coulombs.

$\begin{aligned} & \mathrm{R}_{\text {eq }}=\frac{4}{3} \Omega \\ & \therefore \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}_{\text {eq }}}=\frac{4}{4 / 3}=3 \mathrm{~W} \end{aligned}$

$6\Omega$ is short circuit

$\mathrm{R}_{\mathrm{eq}}=3 \times \frac{1}{3}=1 \Omega$

$\begin{aligned} & \frac{R_v 100}{R_v+100}=\frac{200}{3} \\ & 3 R_v=2 R_v+200 \\ & R_v=200 \end{aligned}$

$I=\frac{8-2}{2+4}=\frac{6}{6}=1 \mathrm{~A}$

Applying Kirchhoff from C to B

$\begin{aligned} & V_C-2-4 \times 1=V_B \\\\ & V_C-V_B=6 \mathrm{~V} \\\\ & =6 \mathrm{~V} \end{aligned}$

Given, $\mathrm{i}_1=0.3 \mathrm{~A}, \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=0.9 \mathrm{~A}$

So, $\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_1 \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$

$\begin{aligned} & \mathrm{i}_2=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A} \\ & \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=\frac{9}{10} \mathrm{~A} \\ & \frac{3}{10}+\frac{2}{5}+\mathrm{i}_3=\frac{9}{10} \\ & \frac{7}{10}+\mathrm{i}_3=\frac{9}{10} \\ & \mathrm{i}_3=0.2 \mathrm{~A} \\ & \mathrm{So}, \mathrm{i}_3 \times \mathrm{R}_1=6 \mathrm{~V} \\ & (0.2) \mathrm{R}_1=6 \\ & \mathrm{R}_1=\frac{6}{0.2}=30 \Omega \end{aligned}$