Physical Sciences Grade 12: Electric Circuits Study Guide

Question 1: Experiment on Emf and Internal Resistance

• Experimental Overview

  • Learners conducted an experiment to determine the electromotive force ($\epsilon$) and the internal resistance ($r$) of a battery.

  • Circuit Components:

    • A battery.

    • A rheostat (variable resistor).

    • A low-resistance ammeter (to measure current $I$).

    • A high-resistance voltmeter (to measure potential difference $V$).

  • Data Recorded:

    • Reading on Voltmeter ($V$): $2\,V, 3\,V, 4\,V, 5\,V, 6\,V$

    • Reading on Ammeter ($A$): $0.58\,A, 0.46\,A, 0.36\,A, 0.24\,A, 0.14\,A$

• Experimental Requirements

  • Constant Factor: The temperature of the battery and the conductors must be kept constant during the experiment to ensure resistance values remain stable.

  • Graphing: Plotting $V$ (Volts) on the vertical axis and $I$ (Amperes) on the horizontal axis yields a straight line with a negative gradient.

  • Determining Emf ($\epsilon$): The emf is the value of the potential difference when the current is zero. On the graph of $V$ vs $I$, this is the y-intercept (vertical axis intercept).

  • Determining Internal Resistance ($r$): Internal resistance is calculated as the negative of the gradient (slope) of the $V$ vs $I$ graph.

    • Calculation: $r = -\text{gradient} = -\frac{\Delta V}{\Delta I}$

• Device X and Y Circuit Calculation (Switch $S_1$ closed, $S_1$ open)

  • Given:

    • Battery Emf: $\epsilon = 24\,V$

    • Device X Rating: $20\,V, 100\,W$

    • Device Y Power Rating: $150\,W$

  • Current in X ($I_X$):

    • $P = V \times I$

    • $100 = 20 \times I_X$

    • $I_X = \frac{100}{20} = 5\,A$

  • Resistance of Y ($R_Y$):

    • Since X and Y are in parallel and functioning at rated values, the voltage across Y is ALSO $20\,V$.

    • $P = \frac{V^2}{R}$

    • $150 = \frac{20^2}{R_Y}$

    • $R_Y = \frac{400}{150} = 2.67\,\Omega$

  • Internal Resistance ($r$):

    • Total current ($I_{total}$) = $I_X + I_Y$

    • $I_Y = \frac{150}{20} = 7.5\,A$

    • $I_{total} = 5 + 7.5 = 12.5\,A$

    • Voltage drop across internal resistance: $v_{int} = \epsilon - V_{ext} = 24 - 20 = 4\,V$

    • $r = \frac{v_{int}}{I_{total}} = \frac{4}{12.5} = 0.32\,\Omega$

• Device Z Functionality

  • Identification: Device Z must be a voltmeter or a component with infinitely high resistance.

  • Explanation: If Z has high resistance, it will not draw current, allowing the total resistance and current in the rest of the circuit to remain unchanged when switch $S_2$ is closed. This ensures X and Y continue to operate at their rated values.

Question 2: Wire Selection and Complex Circuits

• Wire Selection for Electric Heater

  • Fair Test Factors: To compare wires A and B fairly, learners should consider the length of the wires and the thickness (cross-sectional area) of the wires.

  • Suitability Calculation:

    • Gradient of $V$ vs $I$ graph represents resistance ($R$).

    • Wire A: $R_A = \frac{10.0}{0.5} = 20\,\Omega$ (using points from the graph).

    • Wire B: $R_B = \frac{10.0}{1.0} = 10\,\Omega$.

    • For a heater (connected to a constant voltage source like mains), power $P = \frac{V^2}{R}$. Therefore, the wire with the lower resistance (Wire B) will produce more heat/power and is more suitable.

• Circuit Analysis (2.2)

  • Given: $\epsilon = 9\,V$, Ammeter $A = 0.2\,A$ through one branch of an 11-11 parallel pair.

  • Current through $5.5\,\Omega$ Resistor:

    • Since the two $11\,\Omega$ resistors are in parallel, and one has $0.2\,A$, the other also has $0.2\,A$.

    • Total current ($I_{total}$) passing through the series $5.5\,\Omega$ resistor is $0.2 + 0.2 = 0.4\,A$.

  • Internal Resistance Calculation:

    • External Resistance ($R_{ext}$) = $R_{parallel} + R_{series} = (\frac{11 \times 11}{11 + 11}) + 5.5 = 5.5 + 5.5 = 11\,\Omega$

    • $\epsilon = I(R_{ext} + r)$

    • $9 = 0.4(11 + r)$

    • $22.5 = 11 + r \rightarrow r = 11.5\,\Omega$

  • Removal of Resistor effect:

    • If the $5.5\,\Omega$ resistor is removed (which is in series with the whole circuit), the total resistance of the circuit increases (if it were parallel, it would be different, but looking at the diagram it appears to be a series branch).

    • Correction based on diagram: If the $5.5\,\Omega$ resistor is removed from a series position, it creates an open circuit (current is zero). If it is replaced by a wire, resistance decreases and current increases. If it is a parallel branch being removed, total resistance increases and ammeter reading DECREASES because $I_{total} = \frac{\epsilon}{R+r}$.

Question 3: Component Analysis and Parallel Resistors

• General Calculations

  • Ammeter Polarity (Point P): Point P represents the NEGATIVE terminal of the ammeter (connected towards the negative terminal of the cell).

  • Ammeter Reading:

    • $V_{term} = 1.36\,V$ (Voltage across external circuit).

    • $R_{parallel} = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2\,\Omega$

    • $R_{ext} = 4 + 1.2 = 5.2\,\Omega$

    • $I = \frac{V_{ext}}{R_{ext}} = \frac{1.36}{5.2} = 0.26\,A$

  • Internal Resistance ($r$):

    • $V_{int} = \epsilon - V_{ext} = 1.5 - 1.36 = 0.14\,V$

    • $r = \frac{V_{int}}{I} = \frac{0.14}{0.26} = 0.54\,\Omega$

• Impact of Parallel Resistor X:

  • Adding resistor X in parallel to the $3\,\Omega$ resistor DECREASES the total external resistance ($R_{ext}$).

  • Because $I = \frac{\epsilon}{R_{ext} + r}$, the total current in the circuit INCREASES.

  • Consequently, the ammeter reading will INCREASE.

Question 4: Ohm's Law and Power

• Definitions

  • Ohm's Law: The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature.

• Circuit Calculations

  • Reading on $A_1$:

    • Current through $8\,\Omega$ is $0.5\,A$.

    • Voltage across parallel branch: $V = I \times R = 0.5 \times 8 = 4\,V$.

    • Current through $16\,\Omega$ branch: $I_{16} = \frac{4}{16} = 0.25\,A$.

    • Total current ($A_1$) = $0.5 + 0.25 = 0.75\,A$.

  • Reading on $A_2$ (Current through Device R delivering $12\,W$):

    • The current through $A_2$ is the same as the current through R.

    • Note: Without knowing R or the total EMF, we use energy logic. If R is in series with the parallel bank ($4\,V$) and the $20\,\Omega$ resistor ($0.75 \times 20 = 15\,V$), the external voltage depends on $\epsilon$.

    • Alternative: If the power is given as $12\,W$, and we find current $I_{total} = 0.75\,A$, then $P = I^2 R \rightarrow 12 = (0.75)^2 \times R \rightarrow R = \frac{12}{0.5625} = 21.33\,\Omega$. The ammeter $A_2$ reading is the total current $0.75\,A$.

  • Voltmeter Reading (Switch Open): The voltmeter reads the Emf ($\epsilon$).

    • $V_{ext} = V_{parallel} + V_{20} + V_R = 4 + 15 + (I \times R_{device}) = 19 + (0.75 \times 21.33) = 19 + 16 = 35\,V$.

    • $V_{int} = I \times r = 0.75 \times 1 = 0.75\,V$.

    • $\epsilon = 35 + 0.75 = 35.75\,V$.

Question 5: Voltmeter Readings and Circuit Changes

• Calculations

  • Voltmeter Reading $V_1$:

    • Current through top branch ($4\,\Omega$ and $8\,\Omega$) is $0.2\,A$.

    • $V = I \times R = 0.2 \times 8 = 1.6\,V$.

  • Total Current:

    • Voltage across parallel branch = $0.2 \times (4 + 8) = 2.4\,V$.

    • Current in bottom $2\,\Omega$ resistor: $I = \frac{2.4}{2} = 1.2\,A$.

    • $I_{total} = 0.2 + 1.2 = 1.4\,A$.

  • Emf ($\epsilon$):

    • $r = 0.5\,\Omega$.

    • $\epsilon = V_{ext} + I_{total}r = 2.4 + (1.4 \times 0.5) = 2.4 + 0.7 = 3.1\,V$.

• Circuit Modification: Removing the $2\,\Omega$ resistor.

  • Answer: Reading on voltmeter ($V_1$) will INCREASE.

  • Reason: Removing a parallel resistor increases the total external resistance. This decreases the total current ($I_{total}$) supplied by the battery. Consequently, the internal voltage drop ($Ir$) decreases, so the external voltage ($V_{lost} = \epsilon - Ir$) increases. More current will then flow through the remaining top branch, increasing the voltage across the $8\,\Omega$ resistor.

Question 6: Light Bulbs and Internal Resistance

• Parallel Bulbs (Negligible Internal Resistance)

  • Bulbs: A ($4\,W$), B ($6\,W$), C ($10\,W$). Source: $12\,V$.

  • Resistance of 4 W bulb: $P = \frac{V^2}{R} \rightarrow 4 = \frac{12^2}{R} \rightarrow R = \frac{144}{4} = 36\,\Omega$.

  • Equivalent Resistance Change: If bulb B burns out, it is removed from the parallel circuit. Total resistance INCREASES.

  • Power of 10 W Bulb: REMAINS THE SAME. Because there is no internal resistance, the voltage across bulb C remains $12\,V$, so its power dissipation ($P = \frac{V^2}{R}$) does not change.

• Internal Resistance Calculation (6.2)

  • $\epsilon = 6\,V$ (Voltmeter reading when open).

  • $V_{ext} = 5\,V$ (Voltmeter reading when closed across $6\,\Omega$ resistor).

  • Current $I = \frac{V}{R} = \frac{5}{6} = 0.83\,A$.

  • $V_{int} = \epsilon - V = 6 - 5 = 1\,V$.

  • $r = \frac{V_{int}}{I} = \frac{1}{0.83} = 1.2\,\Omega$.

• Adding Resistor X Parallel:

  • $V_{new} = 4.5\,V$.

  • $V_{int} = 6 - 4.5 = 1.5\,V$.

  • $I_{total} = \frac{V_{int}}{r} = \frac{1.5}{1.2} = 1.25\,A$.

  • $R_{ext} = \frac{V_{new}}{I_{total}} = \frac{4.5}{1.25} = 3.6\,\Omega$.

  • $\frac{1}{R_{ext}} = \frac{1}{6} + \frac{1}{X} \rightarrow \frac{1}{3.6} = \frac{1}{6} + \frac{1}{X}$.

  • $\frac{1}{X} = 0.2778 - 0.1667 = 0.1111 \rightarrow X = 9\,\Omega$.

Question 7: Electric Motor and Resistance

• Circuit Properties

  • $\epsilon = 12\,V, r = 0.2\,\Omega$.

  • Switch S Open: Voltmeter across points a and b reads $12\,V$ (Emf).

  • Switch S Closed: Voltmeter across points c and d (parallel branch) reads the terminal voltage minus any series drops. Given $V_{term} = 11.7\,V$.

• Calculations

  • Current in Battery ($I$):

    • $V_{int} = \epsilon - V_{term} = 12 - 11.7 = 0.3\,V$.

    • $I = \frac{V_{int}}{r} = \frac{0.3}{0.2} = 1.5\,A$.

  • Effective Resistance of Parallel Branch:

    • The current splits between resistors $10\,\Omega$ and $15\,\Omega$.

    • $R_p = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\,\Omega$.

  • Resistance of Resistor R:

    • Total external resistance $R_{ext} = \frac{V_{term}}{I} = \frac{11.7}{1.5} = 7.8\,\Omega$.

    • Since R is in series with the parallel branch: $R_{ext} = R_p + R$

    • $7.8 = 6 + R \rightarrow R = 1.8\,\Omega$.

• Electric Motor Analysis

  • Motor Rating X (Watts):

    • Power of motor ($P$) used to lift mass = $F \times v$.

    • $F = mg = 0.35 \times 9.8 = 3.43\,N$.

    • $P = 3.43 \times 0.4 = 1.372\,W$.

    • Therefore, $X = 1.372\,W$.

  • Resistance of Resistor T:

    • Motor Voltage $V_M = 3\,V$ (rated).

    • Motor Current $I_M = \frac{P}{V_M} = \frac{1.372}{3} = 0.457\,A$.

    • Since the motor and resistor T are in series with the battery:

    • $\epsilon = V_M + V_T + I_{total}r$

    • $12 = 3 + (0.457 \times T) + (0.457 \times 0.2)$

    • $12 = 3 + 0.457T + 0.0914$

    • $8.9086 = 0.457T \rightarrow T = 19.5\,\Omega$.