Solution

$$ \begin{eqnarray} I = \frac{dq}{dt} &=& \frac{\Delta q}{\Delta t} \\ &=& \frac{~q~ C}{~t~ s} \\ &=& [[return sf_latex(namespace_resistance.i)]] C/s \\ &=& [[return sf_latex(namespace_resistance.i)]] A \end{eqnarray} $$

Solution

Current density: $$ \begin{eqnarray} j &=& \frac{I}{A} \\ &=& \frac{I}{\pi r^2} \\ &=& \frac{~current~ \times 10^{-3}A}{\pi (~radius~ \times 10^{-3})^2} \\ &=& [[return sf_latex(namespace_resistance.current_density)]] A/m^2 \end{eqnarray} $$

Drift velocity: $$ \begin{eqnarray} I &=& n q v_d A \\ \Rightarrow j &=& \frac{I}{A} = n q v_d \\ \Rightarrow v_d &=& \frac{j}{n q} \\ \Rightarrow |v_d| &=& \frac{j}{n |q|} &\text{ (looking for magnitude only)} \\ &=& \frac{j}{n e} &\text{ (charge carriers in a metal are electrons, with $q = -e$)} \\ &=& \frac{[[return sf_latex(namespace_resistance.current_density)]] A/m^2}{([[return sf_latex(namespace_resistance.density)]] m^{-3}) (1.6\times 10^{-19}C)} \\ &=& [[return sf_latex(namespace_resistance.drift_velocity)]] m/s \end{eqnarray} $$

Solution

Find $R$

$$ \begin{eqnarray} R &=& \frac{V_1}{I_1} \\ &=& \frac{~v_1~ V}{~i_1~ A} \\ &=& [[return sf_latex(namespace_resistance.r)]] \Omega \end{eqnarray} $$

Find $V_2$

$R$ stays the same, so we have: $$ \begin{eqnarray} V_2 &=& I_2 R \\ &=& (~i_2~ A)([[return sf_latex(namespace_resistance.r)]] \Omega) \\ &=& [[return sf_latex(namespace_resistance.v_2)]] V \end{eqnarray} $$

Solution

$$ \begin{eqnarray} R &=& \rho \frac{l}{A} \\ &=& \rho \frac{l}{\pi r^2} \\ &=& ([[return sf_latex(namespace_resistance.resistivity)]] \Omega m) \frac{~l~ \times 0.01m}{\pi (~radius~ \times 10^{-3}m)^2} \\ &=& [[return sf_latex(namespace_resistance.resistance)]] \Omega \end{eqnarray} $$

Solution

Potential difference

$$ \begin{eqnarray} V &=& I R \\ &=& (~i~ A)(~r~ \Omega) \\ &=& [[return sf_latex(namespace_resistance.v)]] V \end{eqnarray} $$

Power

$$ \begin{eqnarray} P &=& I^2 R \\ &=& (~i~ A)^2(~r~ \Omega) \\ &=& [[return sf_latex(namespace_resistance.p)]] W \end{eqnarray} $$

$P = [[return sf_latex(namespace_resistance.p)]] W = [[return sf_latex(namespace_resistance.p)]] J/s$. It means that $[[return sf_latex(namespace_resistance.p)]] J$ is being released by the resistor every second.

Alternatively: $$ \begin{eqnarray} P &=& I V \\ &=& (~i~ A)([[return sf_latex(namespace_resistance.v)]] V) \\ &=& [[return sf_latex(namespace_resistance.p)]] W \end{eqnarray} $$

Solution

$$ \begin{eqnarray} V_{apparent} &=& \mathcal{E} - Ir \\ &=& ~emf~ V - (~i~ A)(~r~ \Omega) \\ &=& [[return sf_latex(namespace_resistance.v_apparent)]] V \end{eqnarray} $$ For $V_{apparent} = [[return ~v_percentage~/100]] \mathcal{E} = [[return sf_latex(~v_percentage~/100 * ~emf~)]] V$, we simply solve for $I$: $$ \begin{eqnarray} V_{apparent} &=& \mathcal{E} - Ir \\ \Rightarrow [[return ~v_percentage~/100]] \mathcal{E} &=& \mathcal{E} - Ir \\ \Rightarrow Ir &=& (1 - [[return ~v_percentage~/100]]) \mathcal{E} \\ \Rightarrow I &=& \frac{[[return (100 - ~v_percentage~)/100]] \mathcal{E}}{r} \\ &=& \frac{([[return (100 - ~v_percentage~)/100]]) (~emf~ V)}{~r~ \Omega} \\ &=& [[return sf_latex(namespace_resistance.current_required)]] A \end{eqnarray} $$

Solution

Solution

By Kirchhoff's loop rule, the total change in potential must be zero, therefore: $$ \begin{eqnarray} 0 &=& \sum_{loop} \Delta V \\ &=& \Delta V_{ba} + \Delta V_{cb} + \Delta V_{dc} \\ \Rightarrow \Delta V_{dc} &=& -\Delta V_{ba} - \Delta V_{cb} \\ &=& -([[return namespace_resistance.v_ba]] V) - ([[return namespace_resistance.v_cb]] V) \\ &=& [[return namespace_resistance.v_dc]] V \end{eqnarray} $$

By Kirchhoff's loop rule, going from $a$ to $d$ basically completes a whole loop, so the total change in potential must be zero: $$ \Delta V_{da} = 0V $$ One can also verify with our previous numbers that: $$ \Delta V_{da} = \Delta V_{dc} + \Delta V_{cb} + \Delta V_{ba} = 0V $$

Solution

Loop 1

$$ \begin{eqnarray} 0 &=& -V_1 + V_3 - I_3 R_3 + I_1 R_1 \tag{1} \end{eqnarray} $$

Loop 2

$$ \begin{eqnarray} 0 &=& V_2 - I_2 R_2 + I_3 R_3 - V_3 \\ &=& V_2 - V_3 - I_2 R_2 + I_3 R_3 \tag{2} \end{eqnarray} $$

Junction rule at $a$

By conservation of current, we have: $$ I_1 + I_2 + I_3 = 0 \tag{3} $$ Together we have three equations for the three unknows ($I_1$, $I_2$ and $I_3$). Solve them any way you like. I will demonstrate one possible approach below.

From eqn(3), we get $I_1 = - I_2 - I_3$. Put into eqn(1): $$ \begin{eqnarray} 0 &=& -V_1 + V_3 - I_3 R_3 + (- I_2 - I_3) R_1 \\ &=& -V_1 + V_3 - I_2 R_1 -I_3 (R_1+R_3) \end{eqnarray} $$ This gives us: $$ \begin{eqnarray} \left\{ % comment a closing brace to remove the error warning } \begin{array}{rll} 0 &= -V_1 + V_3 - I_2 R_1 -I_3 (R_1+R_3) &\text{(1')} \\ 0 &= V_2 - V_3 - I_2 R_2 + I_3 R_3 &\text{(2)} \end{array} \right. \end{eqnarray} $$

Multiply eqn(1') by $R_2$ and eqn(2) by $R_1$: $$ \begin{eqnarray} \left\{ % comment a closing brace to remove the error warning } \begin{array}{rll} 0 &= (-V_1 + V_3)R_2 - I_2 R_1 R_2 -I_3 R_2 (R_1+R_3) & \text{(1'')} \\ 0 &= (V_2 - V_3)R_1 - I_2 R_1 R_2 + I_3 R_1 R_3 & \text{(2')} \end{array} \right. \end{eqnarray} $$

Subtracting eqn(2') from eqn(1'') gives: $$ \begin{eqnarray} 0 &=& (-V_1 + V_3)R_2 - (V_2 - V_3)R_1 - I_3 R_2 (R_1+R_3) - I_3 R_1 R_3 \\ \Rightarrow I_3 (R_2 (R_1+R_3) + R_1 R_3) &=& (-V_1 + V_3)R_2 - (V_2 - V_3)R_1 \\ \Rightarrow I_3 &=& \frac{(-V_1 + V_3)R_2 - (V_2 - V_3)R_1}{R_2 (R_1+R_3) + R_1 R_3} \\ &=& [[return sf_latex(namespace_resistance.i_3)]] A \end{eqnarray} $$

Put $I_3$ back into eqn(2): $$ \begin{eqnarray} 0 &=& V_2 - V_3 - I_2 R_2 + I_3 R_3 \\ \Rightarrow I_2 R_2 &=& V_2 - V_3 + I_3 R_3 \\ \Rightarrow I_2 &=& \frac{V_2 - V_3 + I_3 R_3}{R_2} \\ &=& [[return sf_latex(namespace_resistance.i_2)]] A \end{eqnarray} $$

Finally, eqn(3) gives: $$ \begin{eqnarray} I_1 &=& -I_2 - I_3 \\ &=& [[return sf_latex(namespace_resistance.i_1)]] A \end{eqnarray} $$

Optional technique - matrix solution

Solving the above equations by hand is not difficult but quite tedious. If you know how to use matrices on a calculator, you could also use the method below.

$$ \begin{eqnarray} 0 &=& -V_1 + V_3 - I_3 R_3 + I_1 R_1 \tag{1} \end{eqnarray} $$ $$ \begin{eqnarray} 0 &=& V_2 - I_2 R_2 + I_3 R_3 - V_3 \\ &=& V_2 - V_3 - I_2 R_2 + I_3 R_3 \tag{2} \end{eqnarray} $$

The three equations (1), (2) and (3) above can be written as: $$ \begin{eqnarray} \left\{ % comment a closing brace to remove the error warning } \begin{array}{rl} R_1 I_1 + 0 I_2 - R_3 I_3 &= V_1 - V_3 \\ 0 I_1 - R_2 I_2 + R_3 I_3 &= -V_2 + V_3 \\ I_1 + I_2 + I_3 &= 0 \end{array} \right. \end{eqnarray} $$ These equations can be written as: $$ \left(\begin{array}{ccc}R_1 & 0 & -R_3 \\0 & -R_2 & R_3 \\1 & 1 & 1\end{array}\right) \left(\begin{array}{c}I_1 \\I_2 \\I_3\end{array}\right) = \left(\begin{array}{c}V_1 - V_3 \\-V_2 + V_3 \\0\end{array}\right) $$

Calling the matrices and vectors above ${\cal R}, {\cal I}$, and ${\cal V}$, we can write the equation as: $$ \begin{eqnarray} {\cal R} {\cal I} &=& {\cal V} \\ \Rightarrow {\cal I} &=& {\cal R}^{-1} {\cal V} \end{eqnarray} $$ Use your calculator to find the inverse matrix ${\cal R}^{-1}$ and multiply to ${\cal V}$ then gives: $$ \begin{eqnarray} \Rightarrow {\cal I} = \left(\begin{array}{c}I_1 \\I_2 \\I_3\end{array}\right) &=& \left(\begin{array}{c}[[return sf_latex(namespace_resistance.i_1)]] A \\[[return sf_latex(namespace_resistance.i_2)]] A \\[[return sf_latex(namespace_resistance.i_3)]] A \end{array}\right) \\ \end{eqnarray} $$

Solution

Loop 1

Applying Kirchhoff's loop rule to the loop on the left, starting from the left of the battery: $$ \begin{eqnarray} 0 &=& \sum \Delta V \\ &=& -V_1 - I_3 R_3 + I_1 R_1 \\ &=& - ~v_1~ - ~r_3~ I_3 + ~r_1~ I_1 \\ &=& ~r_1~ I_1 - ~r_3~ I_3 - ~v_1~ \tag{1} \end{eqnarray} $$

Loop 2

Applying Kirchhoff's loop rule to the loop on the right, starting from the left of the battery: $$ \begin{eqnarray} 0 &=& \sum \Delta V \\ &=& +V_2 + I_3 R_3 \\ &=& + ~v_2~ + ~r_3~ I_3 \\ \Rightarrow I_3 &=& -\frac{~v_2~}{~r_3~} A = [[return sf_latex(i_3)]]A \tag{2} \end{eqnarray} $$ Put (2) into (1): $$ \begin{eqnarray} 0 &=& ~r_1~ I_1 - 30 I_3 - 1 \\ &=& ~r_1~ I_1 - 30 (-\frac{~v_2~}{~r_3~}) - 1 \\ &=& ~r_1~ I_1 + ~v_2~ - ~v_1~ \\ &=& ~r_1~ I_1 + [[return sf_latex(~v_2~ - ~v_1~, 0)]] \\ \Rightarrow I_1 &=& -\frac{[[return sf_latex(~v_2~ - ~v_1~, 0)]]}{~r_1~} A = [[return sf_latex(i_1)]] A \end{eqnarray} $$

Junction rule

Apply Kirchhoff's junction rule to point $a$: $$ \begin{eqnarray} 0 &=& \sum I \\ &=& I_1 + I_2 + I_3 \qquad \text{(all + because every $I$ flows INTO $a$)} \\ &=& ([[return sf_latex(i_1)]]A) + I_2 + ([[return sf_latex(i_3)]]A) \\ \Rightarrow I_2 &=& -([[return sf_latex(i_1)]]A) - ([[return sf_latex(i_3)]]A) \\ &=& [[return sf_latex(i_2)]]A \end{eqnarray} $$

Summary

$$ \begin{eqnarray} I_1 &=& [[return sf_latex(i_1)]]A \\ I_2 &=& [[return sf_latex(i_2)]]A \\ I_3 &=& [[return sf_latex(i_3)]]A \end{eqnarray} $$ The negative currents flow opposite to the arrows in the figure.

Solution

Loop 1

Applying Kirchhoff's loop rule to the loop on the left, starting from the left of the battery: $$ \begin{eqnarray} 0 &=& \sum \Delta V \\ &=& - V_1 + V_3 - I_3 R_3 + I_1 R_1 \\ &=& - V_1 + ~v_3~ - ~r_3~ I_3 + ~r_1~ I_1 \\ \Rightarrow V_1 &=& ~r_1~ I_1 - ~r_3~ I_3 + ~v_3~ \tag{1} \end{eqnarray} $$

Loop 2

Applying Kirchhoff's loop rule to the loop on the right, starting from the left of the battery: $$ \begin{eqnarray} 0 &=& \sum \Delta V \\ &=& - I_2 R_2 + I_3 R_3 - V_3 \\ \Rightarrow I_3 R_3 &=& I_2 R_2 + V_3 \\ \Rightarrow I_3 &=& \frac{I_2 R_2 + V_3}{R_3} \\ &=& \frac{(~r_2~) (~i_2~) + ~v_3~}{~r_3~} \\ &=& [[return sf_latex(i_3)]]A \tag{2} \end{eqnarray} $$

Junction rule

Apply Kirchhoff's junction rule to point $a$: $$ \begin{eqnarray} 0 &=& \sum I \\ &=& I_1 + I_2 + I_3 \qquad \text{(all + because every $I$ flows INTO $a$)} \\ \Rightarrow I_1 &=& - I_2 - I_3 \\ &=& -(~i_2~ A) - ([[return sf_latex(i_3)]]A) \\ &=& [[return sf_latex(i_1)]]A \end{eqnarray} $$ Put $I_1 = [[return sf_latex(i_1)]]A$ and $I_3 = [[return sf_latex(i_3)]]A$ into (1): $$ \begin{eqnarray} V_1 &=& ~r_1~ I_1 - ~r_3~ I_3 + ~v_3~ \\ &=& (~r_1~) ([[return sf_latex(i_1)]]) - (~r_3~) ([[return sf_latex(i_3)]]) + ~v_3~ \\ &=& [[return sf_latex(v_1)]] V \end{eqnarray} $$

Summary

$$ \begin{eqnarray} I_1 &=& [[return sf_latex(i_1)]]A \\ I_3 &=& [[return sf_latex(i_3)]]A \\ V_1 &=& [[return sf_latex(v_1)]]V \end{eqnarray} $$ The negative currents flow opposite to the arrows in the figure.

Solution

$R_x$

$$ \begin{eqnarray} R_x &=& \rho \frac{l_x}{A_{yz}} \\ &=& \rho \frac{x}{yz} \\ &=& (~rho~ \times 10^{-8} \Omega m) \frac{~x~ \times 0.01 m}{(~y~ \times 0.01 m) (~z~ \times 0.01 m)} \\ &=& [[return sf_latex(R_x)]] \Omega \end{eqnarray} $$

$R_y$

$$ \begin{eqnarray} R_y &=& \rho \frac{l_y}{A_{xz}} \\ &=& \rho \frac{y}{xz} \\ &=& (~rho~ \times 10^{-8} \Omega m) \frac{~y~ \times 0.01 m}{(~x~ \times 0.01 m) (~z~ \times 0.01 m)} \\ &=& [[return sf_latex(R_y)]] \Omega \end{eqnarray} $$

$R_z$

$$ \begin{eqnarray} R_z &=& \rho \frac{l_z}{A_{xy}} \\ &=& \rho \frac{z}{xy} \\ &=& (~rho~ \times 10^{-8} \Omega m) \frac{~z~ \times 0.01 m}{(~x~ \times 0.01 m) (~y~ \times 0.01 m)} \\ &=& [[return sf_latex(R_z)]] \Omega \end{eqnarray} $$

Solution

$R_{ab}$

For a current passing from $a$ to $b$, the triangle is equivalent to the circuit show to the right. $$ \begin{eqnarray} R_{23} &=& R_2 + R_3 = ~R_2~ + ~R_3~ = [[return sf_latex(R_23, 0)]] \Omega \\ \Rightarrow R_{ab} &=& (R_1^{-1} + R_{23}^{-1})^{-1} \\ &=& (~R_1~ ^{-1} + [[return sf_latex(R_23, 0)]] ^{-1})^{-1} \Omega \\ &=& [[return sf_latex(R_ab)]] \Omega \end{eqnarray} $$

$R_{bc}$

For a current passing from $b$ to $c$: $$ \begin{eqnarray} R_{12} &=& R_1 + R_2 = ~R_1~ + ~R_2~ = [[return sf_latex(R_12, 0)]] \Omega \\ \Rightarrow R_{bc} &=& (R_3^{-1} + R_{12}^{-1})^{-1} \\ &=& (~R_3~ ^{-1} + [[return sf_latex(R_12, 0)]] ^{-1})^{-1} \Omega \\ &=& [[return sf_latex(R_bc)]] \Omega \end{eqnarray} $$

$R_{ac}$

For a current passing from $a$ to $c$: $$ \begin{eqnarray} R_{13} &=& R_1 + R_3 = ~R_1~ + ~R_3~ = [[return sf_latex(R_13, 0)]] \Omega \\ \Rightarrow R_{ac} &=& (R_2^{-1} + R_{13}^{-1})^{-1} \\ &=& (~R_2~ ^{-1} + [[return sf_latex(R_13, 0)]] ^{-1})^{-1} \Omega \\ &=& [[return sf_latex(R_ac)]] \Omega \end{eqnarray} $$

Solution

When $V_G = 0V$

$\frac{R_X}{R_3}$

There is a current $I_B$ flowing from $A$ to $B$ to $C$, and another current $I_D$ flowing from $A$ to $D$ to $C$.

Applying Kirchhoff's loop rule to $BADB$: $$ \begin{eqnarray} 0 &=& \Delta V_{AB} + \Delta V_{DA} + \Delta V_{BD} \qquad \text{notation: $\Delta V_{AB} = V_A - V_B$} \\ &=& I_B R_3 - I_D R_1 + 0 \\ \Rightarrow I_B R_3 &=& I_D R_1 \tag{1} \end{eqnarray} $$ Applying Kirchhoff's loop rule to $BDCB$: $$ \begin{eqnarray} 0 &=& \Delta V_{DB} + \Delta V_{CD} + \Delta V_{BC} \\ &=& 0 - I_D R_2 + I_B R_X \\ \Rightarrow I_B R_X &=& I_D R_2 \tag{2} \end{eqnarray} $$ Taking the ratio $\frac{(2)}{(1)}$: $$ \begin{eqnarray} \frac{I_B R_X}{I_B R_3} &=& \frac{I_D R_2}{I_D R_1} \\ \Rightarrow \frac{R_X}{R_3} &=& \frac{R_2}{R_1} \end{eqnarray} $$

---

Numerical value of $R_X$

$$ \begin{eqnarray} \frac{R_X}{R_3} &=& \frac{R_2}{R_1} \\ \Rightarrow R_X &=& \frac{R_2}{R_1} R_3 \\ &=& \frac{~R_2~}{~R_1~} (~R_3~ \Omega) \\ &=& [[return sf_latex(R_X_1)]] \Omega \end{eqnarray} $$

When $V_G \neq 0V$

Two equations

If $I_G \approx 0A$, then the current flowing through $R_1$ and $R_2$ are the same, which we will call $I_D$. Similarly, the current through $R_3$ and $R_X$ are the same, which we will call $I_B$.

Applying Kirchhoff's loop rule to $BADB$: $$ \begin{eqnarray} 0 &=& \Delta V_{AB} + \Delta V_{DA} + \Delta V_{BD} \qquad \text{notation: $\Delta V_{AB} = V_A - V_B$} \\ &=& I_B R_3 - I_D R_1 - V_G \\ \Rightarrow I_B R_3 &=& I_D R_1 + V_G \tag{3} \end{eqnarray} $$ Applying Kirchhoff's loop rule to $BDCB$: $$ \begin{eqnarray} 0 &=& \Delta V_{DB} + \Delta V_{CD} + \Delta V_{BC} \\ &=& V_G - I_D R_2 + I_B R_X \\ \Rightarrow I_B R_X &=& I_D R_2 - V_G \tag{4} \end{eqnarray} $$

---

$I_D$

Applying Kirchhoff's loop rule to the battery and $R_1$, $R_2$: $$ \begin{eqnarray} 0 &=& {\cal E} - I_D R_1 - I_D R_2 \\ \Rightarrow I_D &=& \frac{\cal E}{R_1 + R_2} \tag{5} \end{eqnarray} $$

---

$\frac{R_X}{R_3}$

Taking the ratio $\frac{(4)}{(3)}$: $$ \begin{eqnarray} \frac{I_B R_X}{I_B R_3} &=& \frac{I_D R_2 - V_G}{I_D R_1 + V_G} \\ \Rightarrow \frac{R_X}{R_3} &=& \frac{\frac{\cal E}{R_1 + R_2} R_2 - V_G}{\frac{\cal E}{R_1 + R_2} R_1 + V_G} \qquad \text{ substitute in (5)} \\ &=& \frac{\frac{R_2}{R_1 + R_2} {\cal E} - V_G}{\frac{R_1}{R_1 + R_2} {\cal E} + V_G} \\ &=& \frac{R_2 {\cal E} - (R_1 + R_2) V_G}{R_1 {\cal E} + (R_1 + R_2) V_G} \end{eqnarray} $$

---

Numerical value of $R_X$

$$ \begin{eqnarray} \frac{R_X}{R_3} &=& \frac{R_2 {\cal E} - (R_1 + R_2) V_G}{R_1 {\cal E} + (R_1 + R_2) V_G} \\ \Rightarrow R_X &=& \frac{R_2 {\cal E} - (R_1 + R_2) V_G}{R_1 {\cal E} + (R_1 + R_2) V_G} R_3 \\ &=& \frac{~R_2~ (~emf~) - (~R_1~ + ~R_2~) (~V_G~)}{~R_1~ (~emf~) + (~R_1~ + ~R_2~) (~V_G~)} (~R_3~ \Omega) \\ &=& [[return sf_latex(R_X_2)]] \Omega \end{eqnarray} $$