Direct Current Circuits: Graphical Resistance Analysis and Branch Potential Calculations

Fundamental Principles of Electric Current, Resistance, and Ohm's Law

• Ohm's Law defines the fundamental linear relationship governing ideal conductors, stating that the electric current ($I$) flowing through a conductor between two points is directly proportional to the potential difference ($V$) across the two points, provided physical conditions such as temperature remain constant:   $V = I \times R$

• Electrical resistance ($R$) measures the opposition to the flow of electric charge and is calculated as the ratio of potential difference to current intensity:   $R = \frac{V}{I}$

• The potential difference ($\Delta V$) across a resistor branch represents the drop in electrical potential as conventional current traverses the component from higher electric potential to lower electric potential:   $\Delta V = V_{\text{initial}} - V_{\text{final}} = I \times R$

• In passive resistive circuit branches, electric current always flows in the direction of decreasing electric potential. If a conductor connects two terminals at potentials $V_A$ and $V_B$, and the conventional current flows from point $A$ to point $B$, then $V_A > V_B$, and the net potential drop is:   $V_A - V_B = I \times R$

Graphical Analysis of Current-Voltage Characteristics

• In a Cartesian coordinate system plotting current intensity ($I$) in amperes ($\text{A}$) on the vertical axis against potential difference ($V$) in volts ($\text{V}$) on the horizontal axis, the resulting characteristic curves for ohmic conductors are straight lines passing through the origin $(0, 0)$.

• The mathematical slope of any line on an $I-V$ coordinate plane represents the electrical conductance ($G$), which is the mathematical reciprocal of electrical resistance ($R$):   $\text{Slope} = \frac{\Delta I}{\Delta V} = \frac{1}{R}$

• Resistance ($R$) is therefore inversely proportional to the slope of the line:   $R = \frac{1}{\text{Slope}}$

• When comparing two linear conductors $x$ and $y$ plotted on identical axes:

  • Conductor $x$ displays a steeper line relative to the horizontal axis than conductor $y$, indicating:     $\text{Slope}_x > \text{Slope}_y$
  • Because resistance is the reciprocal of the slope, the resistance of conductor $x$ is smaller than that of conductor $y$:     $R_x < R_y$

• The ratio of the resistances of conductors $x$ and $y$ ($\frac{R_x}{R_y}$) is given by the inverse ratio of their graphical slopes:   $\frac{R_x}{R_y} = \frac{\text{Slope}_y}{\text{Slope}_x}$

• Analytical determination of the resistance ratio from the grid:

  • Iso-potential analysis (constant $V$): Projecting a vertical line at a fixed potential difference $V$, the current $I_x$ passing through conductor $x$ is twice the current $I_y$ passing through conductor $y$ ($I_x = 2 \times I_y$).
  • Substituting these values into Ohm's Law:     $R_x = \frac{V}{I_x} = \frac{V}{2 \times I_y}$$R_y = \frac{V}{I_y}$
  • Taking the direct ratio yields:     $\frac{R_x}{R_y} = \frac{\frac{V}{2 \times I_y}}{\frac{V}{I_y}} = \frac{1}{2}$

• Iso-current analysis (constant $I$):

  • Projecting a horizontal line at a fixed current intensity $I$, the potential difference required across conductor $y$ ($V_y$) is twice the potential difference required across conductor $x$ ($V_x$), such that $V_y = 2 \times V_x$.
  • Evaluating the resistances:     $R_x = \frac{V_x}{I}$$R_y = \frac{V_y}{I} = \frac{2 \times V_x}{I}$
  • Taking the ratio yields:     $\frac{R_x}{R_y} = \frac{V_x}{2 \times V_x} = \frac{1}{2}$

• Conclusion:

  • The ratio between the resistance of conductor $x$ and conductor $y$ is equal to $\frac{1}{2}$.

Electric Potential Calculations in Resistive Branches

• Given two independent conductor branches carrying currents $I_1$ and $I_2$, where the ratio of current intensities is:   $\frac{I_1}{I_2} = \frac{1}{3}$

• First Conductor Parameters:

  • Left terminal electric potential: $V_{\text{left}, 1} = +5\,\text{V}$
  • Right terminal electric potential: $V_{\text{right}, 1} = -7\,\text{V}$
  • Branch resistance: $R_1 = 1.5\,R$
  • Current flow direction: Directed from left to right ($I_1$)

• Second Conductor Parameters:

  • Left terminal electric potential: Unknown potential at point $X$, denoted $V_X$
  • Right terminal electric potential: $V_{\text{right}, 2} = +22\,\text{V}$
  • Branch resistance: $R_2 = R$
  • Current flow direction: Directed from right to left ($I_2$)

• Multiple-choice options provided for the potential of point $X$ ($V_X$):

  • Option أ: $-2\,\text{V}$
  • Option ب: $2\,\text{V}$
  • Option ج: $5\,\text{V}$
  • Option د: $-5\,\text{V}$

• Step-by-Step Derivation and Solution:

  • Step 1: Compute the potential difference across the first branch ($\Delta V_1$):
  • Current $I_1$ flows from the higher potential ($+5\,\text{V}$) to the lower potential ($-7\,\text{V}$).     $\Delta V_1 = V_{\text{left}, 1} - V_{\text{right}, 1} = (+5\,\text{V}) - (-7\,\text{V}) = 5 + 7 = 12\,\text{V}$
  • Step 2: Formulate the current $I_1$ using Ohm's Law:     $I_1 = \frac{\Delta V_1}{R_1} = \frac{12\,\text{V}}{1.5\,R} = \frac{8}{R}$
  • Step 3: Determine the current $I_2$ through the given ratio:
  • Given $\frac{I_1}{I_2} = \frac{1}{3}$, solving for $I_2$ gives:       $I_2 = 3 \times I_1$
  • Substituting the value of $I_1$ into the expression:       $I_2 = 3 \times \left(\frac{8}{R}\right) = \frac{24}{R}$
  • Step 4: Determine the potential difference across the second branch ($\Delta V_2$):
  • Applying Ohm's Law across the second resistor ($R_2 = R$):       $\Delta V_2 = I_2 \times R_2 = \left(\frac{24}{R}\right) \times R = 24\,\text{V}$
  • Step 5: Express $\Delta V_2$ in terms of terminal potentials to isolate $V_X$:
  • Because the current arrow for $I_2$ points from right to left (from the $+22\,\text{V}$ terminal toward point $X$), the right terminal is at the higher electric potential:       $\Delta V_2 = V_{\text{right}, 2} - V_X$
  • Substituting known quantities into the equation:       $24\,\text{V} = 22\,\text{V} - V_X$
  • Step 6: Solve for $V_X$:     $V_X = 22\,\text{V} - 24\,\text{V} = -2\,\text{V}$

• The resulting electric potential of point $X$ is $-2\,\text{V}$, corresponding to Choice أ.

Analytical Summary and Key Takeaways

• Graphical Axis Identification:

  • It is critical to differentiate between $I-V$ graphs (current on the vertical axis) and $V-I$ graphs (voltage on the vertical axis).
  • In an $I-V$ graph, $\text{Slope} = \frac{1}{R}$. The conductor with the larger slope has the lower resistance.
  • In a $V-I$ graph, $\text{Slope} = R$. The conductor with the larger slope has the higher resistance.
  • Misidentifying the axes leads to the inverted resistance ratio ($2$ instead of $\frac{1}{2}$).

• Potential Arithmetic with Negative Numbers:

  • Subtracting a negative potential value increases the net potential difference: $\Delta V = 5 - (-7) = 12\,\text{V}$.
  • When electric potential drops below zero across a component, the final potential is strictly negative: $V_{\text{final}} = V_{\text{initial}} - \Delta V = 22 - 24 = -2\,\text{V}$.