Electrical Circuits: Resistance, Power, and Ohm's Law Study Guide

Electrical Power and Bulb Ratings

• Problem Statement and Rating Evaluation:

  • An electric bulb is provided with a specific rating of $220\,V$ and $100\,W$.
  • The rating indicates that the bulb consumes $100\,W$ of power when operated at a potential difference of $220\,V$.

• Operating at Reduced Voltage:

  • The problem asks for the power consumed when the bulb is operated on a reduced potential difference of $110\,V$.
  • To solve this, the resistance ($R$) of the bulb is first calculated using the standard power formula:       $P = \frac{V^2}{R}$
  • Rearranging to find Resistance:       $R = \frac{V^2}{P}$$R = \frac{220^2}{100}$$R = \frac{48400}{100} = 484\,\Omega$
  • Calculating the new power ($P'$) at $110\,V$:       $P' = \frac{(V')^2}{R}$$P' = \frac{110^2}{484}$$P' = \frac{12100}{484} = 25\,W$

• Conclusion and Options:

  • The power consumed at $110\,V$ is calculated to be $25\,W$.
  • Provided Multiple Choice Options:
    • (a) $100\,W$
    • (b) $75\,W$
    • (c) (Inferred from text as potential error, likely $50\,W$)
    • (d) $25\,W$ (This is the correct answer).

Series and Parallel Heat Dissipation

• Experimental setup:

  • Two conducting wires of the same material are used.
  • The wires are of equal lengths and equal diameters, which implies their individual resistances ($R$) are identical ($R_1 = R_2 = R$).

• Case 1: Series Connection:

  • The wires are first connected in series across a potential difference ($V$).
  • The equivalent resistance ($R_s$) is:       $R_s = R + R = 2R$
  • The heat produced ($H_s$) in time ($t$) is given by:       $H_s = \frac{V^2}{R_s} \times t = \frac{V^2}{2R} \times t$

• Case 2: Parallel Connection:

  • The wires are then connected in parallel across the same potential difference ($V$).
  • The equivalent resistance ($R_p$) is:       $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$
  • The heat produced ($H_p$) in time ($t$) is given by:       $H_p = \frac{V^2}{R_p} \times t = \frac{V^2}{R/2} \times t = \frac{2V^2}{R} \times t$

• Ratio of Heat Produced:

  • The ratio of heat produced in series to parallel combinations is:       $\frac{H_s}{H_p} = \frac{V^2 t / 2R}{2V^2 t / R}$$\frac{H_s}{H_p} = \frac{1}{4}$       Ratio = $1:4$

• Multiple Choice Options:

  • (a) $1:2$
  • (b) $2:1$
  • (c) $1:4$ (Correct)
  • (d) $4:1$

Circuit Measurements: Voltmeter Application

• Connection Method:
  • A voltmeter is an instrument used to measure the potential difference between two points in a circuit.
  • It is always connected in parallel to the component or the two points across which the potential difference is to be measured.
  • This parallel connection ensures that the voltmeter receives the same potential difference as the component without significantly drawing current from the main circuit (due to its high internal resistance).

Wire Resistivity and Geometric Properties

• Copper Wire Variables:

  • Diameter ($d$) = $0.5\,mm = 0.0005\,m$
  • Resistivity ($\rho$) = $1.6 \times 10^{-8}\,\Omega\,m$
  • Desired Resistance ($R$) = $10\,\Omega$

• Calculating Length ($L$):

  • The formula for resistance is:       $R = \rho \frac{L}{A}$
  • Area ($A$) is calculated using the diameter:       $A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$
  • Substituting values to find $L$:       $L = \frac{R \times A}{\rho}$$L = \frac{10 \times 3.14 \times (0.25 \times 10^{-3})^2}{1.6 \times 10^{-8}}$
  • Based on the provided calculations in the transcript:       $L = 122.66\,m$

• Effect of Doubling the Diameter:

  • If the diameter is doubled, the new diameter becomes ($2 \times 0.5\,mm$) = $1\,mm$, which is $0.001\,m$.
  • Relationship between Resistance and Diameter: $R \propto \frac{1}{d^2}$.
  • When diameter ($d$) is doubled, the area quadruples ($2^2 = 4$), causing the resistance to become one-fourth of its original value.
  • Verification with formula:       $R = \rho \frac{L}{A} = 1.6 \times 10^{-8} \times \frac{122.66}{3.14 \times (0.0005)^2 \text{ (original area factor)}}$
  • Transcript Result for new resistance ($R'$) with doubled diameter:       $R' = 2.5\,\Omega$
  • Note: Since the original $R = 10\,\Omega$, and $10 / 4 = 2.5\,\Omega$, the relationship holds that resistance remains one-fourth ($1/4\,R$) when diameter is doubled.

Ohm's Law and V-I Characteristics

• Data Set for Potential Difference ($V$) and Current ($I$):

  • The transcript provides the following experimental values for a resistor:
    • Current ($I$ in amperes): $0.5$, $1.0$, $2.0$, $3.0$, $4.0$
    • Potential Difference ($V$ in volts): $1.6$, $3.4$, $6.7$, $10.2$, $13.2$

• Graphing and Calculation:

  • A graph should be plotted with Potential Difference ($V$) on the y-axis and Current ($I$) on the x-axis.
  • According to Ohm's Law ($V = IR$), the resulting plot is a straight line passing through the origin.
  • The resistance ($R$) of the resistor is calculated by finding the slope of this graph.
  • Slope formula:       $\text{Resistance (R)} = \text{Slope} = \frac{\Delta V}{\Delta I}$
  • This slope represents the constant resistance value for the given component at a constant temperature.