Ohm's Law and Resistivity Worksheet

Fundamental Ohm's Law Calculations

• Ohm's Law Relationship: The relationship between voltage ($V$), current ($I$), and resistance ($R$) is defined by the formula $I = \frac{V}{R}$, which can be rearranged to $V = I \times R$ or $R = \frac{V}{I}$.

• Application 1: Low Voltage/Low Resistance Circuit
    - Problem: Determine the current in a $10\,V$ circuit with a resistance of $2\,\Omega$.
    - Variables Provided:
      - Voltage ($V$): $10\,V$
      - Resistance ($R$): $2\,\Omega$
    - Calculation:
      - $I = \frac{V}{R}$
      - $I = \frac{10}{2}$
    - Result: Current ($I$) = $5\,A$

• Application 2: Standard Voltage Circuit with Moderate Resistance
    - Problem: Determine the current in a $120\,V$ circuit with a resistance of $20\,\Omega$.
    - Variables Provided:
      - Voltage ($V$): $120\,V$
      - Resistance ($R$): $20\,\Omega$
    - Calculation:
      - $I = \frac{V}{R}$
      - $I = \frac{120}{20}$
    - Result: Current ($I$) = $6\,A$

• Application 3: Standard Voltage Circuit with Decreased Resistance
    - Problem: Determine the current in a $120\,V$ circuit with a resistance of $10\,\Omega$.
    - Variables Provided:
      - Voltage ($V$): $120\,V$
      - Resistance ($R$): $10\,\Omega$
    - Calculation:
      - $I = \frac{V}{R}$
      - $I = \frac{120}{10}$
    - Result: Current ($I$) = $12\,A$

• Application 4: Car Battery Headlight Circuit
    - Problem: A $10\,V$ car battery pushes charge through a headlight circuit with a resistance of $10\,\Omega$. Calculate the current.
    - Variables Provided:
      - Voltage ($V$): $10\,V$
      - Resistance ($R$): $10\,\Omega$
    - Calculation:
      - $I = \frac{V}{R}$
      - $I = \frac{10}{10}$
    - Result: Current ($I$) = $1\,A$

• Application 5: Electric Heater Voltage Requirements
    - Problem: An electric heater passes a current of $100\,A$ through a coiled metal wire. If the wire's resistance is $2.5\,\Omega$, what voltage must be applied?
    - Variables Provided:
      - Current ($I$): $100\,A$
      - Resistance ($R$): $2.5\,\Omega$
    - Calculation:
      - $V = I \times R$
      - $V = (100) \times (2.5)$
    - Result: Voltage ($V$) = $250\,V$

Charge, Current, and Time Relationships

• Charge Definition: The relationship between current, charge ($Q$), and time ($t$) is given by $I = \frac{Q}{t}$. This is rearranged to solve for charge as $Q = I \times t$.

• Case Study 1: Flashlight Battery (1.5 V)
    - Problem: A battery with an Electromotive Force (emf) of $1.5\,V$ delivers a current of $0.25\,A$ to a bulb for $64\,sec$. Find the charge that passes through the circuit.
    - Variables Provided:
      - Current ($I$): $0.25\,A$
      - Time ($t$): $64\,s$
      - Voltage (for context): $1.5\,V$
    - Calculation:
      - $Q = I \times t$
      - $Q = (0.25) \times (64)$
    - Result: Charge ($Q$) = $16\,C$

• Case Study 2: Flashlight Battery (3 V)
    - Problem: A battery with a potential of $3\,V$ delivers a current of $0.15\,A$ for $30\,sec$. Find the total charge in Coulombs.
    - Variables Provided:
      - Current ($I$): $0.15\,A$
      - Time ($t$): $30\,s$
      - Voltage (for context): $3\,V$
    - Calculation:
      - $Q = I \times t$
      - $Q = (0.15) \times (30)$
    - Result: Charge ($Q$) = $4.5\,C$

Resistance and Resistivity in Physical Conductors

• Resistance and Geometry: The resistance of a conductor depends on its resistivity ($\rho$), length ($L$), and cross-sectional area ($A$): $R = \frac{\rho \times L}{A}$.

• Potential Difference in a Heavy Gauge Copper Wire
    - Problem: A current of $50\,A$ flows through a copper wire $1.75\,m$ long and $0.1\,m$ in diameter. Given a resistivity ($\rho$) of $1000\,\Omega \cdot m$, find the potential difference ($V$) between the ends.
    - Step 1: Calculate Cross-Sectional Area ($A$):
      - Formula: $A = \frac{\pi \times d^2}{4}$
      - Substitution: $A = \frac{(3.14) \times (0.1)^{2}}{4}$
      - Result: $A = 7.85 \times 10^{-3}\,m^{2}$
    - Step 2: Calculate Resistance ($R$):
      - Formula: $R = \frac{\rho \times L}{A}$
      - Substitution: $R = \frac{(1000) \times (1.75)}{7.85 \times 10^{-3}}$
      - Result: $R = 222929.92\,\Omega$
    - Step 3: Calculate Potential Difference ($V$):
      - Formula: $V = I \times R$
      - Substitution: $V = (50) \times (222929.92)$
      - Result: $V = 11146494.8\,V$

• Determining Resistivity of an Unknown Conductor
    - Problem: A potential difference of $12\,V$ applied to a wire ($5\,m$ long, $0.25\,m$ diameter) results in a current of $15\,A$. Calculate the wire's resistivity ($\rho$).
    - Step 1: Determine Radius ($r$):
      - $r = \frac{d}{2} = \frac{0.25}{2}$
      - Result: $r = 0.125\,m$
    - Step 2: Calculate Cross-Sectional Area ($A$):
      - Formula: $A = \pi \times r^2$
      - Substitution: $A = (3.14) \times (0.125)^2$
      - Rounded Result: $A = 0.05\,m^2$
    - Step 3: Calculate Resistance ($R$):
      - formula: $R = \frac{V}{I}$
      - Substitution: $R = \frac{12}{15}$
      - Result: $R = 0.8\,\Omega$
    - Step 4: Calculate Resistivity ($\rho$):
      - Derived Formula: $\rho = \frac{R \times A}{L}$
      - Substitution: $\rho = \frac{(0.8) \times (0.05)}{5}$
      - Result: $\rho = 0.008\,\Omega \cdot m$