Physical Sciences P1 Physics Grade 12 - Electricity and Circuits Study Notes

Defining Electromotive Force and Fundamental Circuit Principles The electromotive force ($\varepsilon$ or emf) of a battery is defined as the maximum energy supplied by a battery per coulomb of charge ($C$) passing through it. This value represents the total electrical potential energy per unit charge that the battery can provide to the entire circuit, including its own internal components. When a circuit is in an open state (switch S is open), a high-resistance voltmeter connected across the terminals of the battery measures the emf directly because there is no current flowing, and thus no potential drop across the internal resistance. In the context of Question 18.1, the definition is completed by identifying that (a) represents maximum energy and (b) represents unit charge (or coulomb of charge). # Theoretical and Practical Analysis of Internal Resistance and Potential Drop A critical observation in practical circuits is that the reading on a voltmeter connected across a battery decreases when the switch is closed. In Question 17, the voltmeter reading decreases by $1.5\,V$ upon closing switch S. This reduction in terminal potential difference occurs because the battery possesses an internal resistance ($r$). When the switch is closed, a current ($I$) flows through the circuit. As this current passes through the battery, some of the battery's energy is dissipated internally to overcome this resistance. This internal dissipation is known as the "lost volts" ($V_{\text{internal}}$), which is mathematically expressed by the product of the current and the internal resistance: $V_{\text{internal}} = I \times r$. Because this potential is used inside the battery, it is no longer available to the external circuit, causing the terminal potential difference ($V_{\text{external}}$) to be less than the emf ($\varepsilon$). # Detailed Calculations for Question 17 Circuit Network In Question 17, the circuit consists of a battery with internal resistance $r = 0.5\,\Omega$ and unknown emf ($\varepsilon$), connected to a series-parallel network containing resistors $R_1 = 4\,\Omega$, $R_2 = 25\,\Omega$, and $R_3 = 15\,\Omega$. To solve for the ammeter reading (17.3.1), we utilize the given decrease in voltmeter reading, which is the internal potential drop ($V_{\text{internal}} = 1.5\,V$). Using Ohm's Law for the internal component: $V_{\text{internal}} = I \times r$, we substitute the values to find $1.5\,V = I \times 0.5\,\Omega$, which results in a current of $I = 3\,A$. To determine the total external resistance ($R_{\text{ext}}$) as required in 17.3.2, we must first calculate the parallel resistance of $R_2$ and $R_3$. The formula for parallel resistors is $\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3}$, which is $\frac{1}{R_p} = \frac{1}{25\,\Omega} + \frac{1}{15\,\Omega}$. Calculating the common denominator or using the product-over-sum method: $R_p = \frac{25 \times 15}{25 + 15} = \frac{375}{40} = 9.375\,\Omega$. The total external resistance is the sum of the series resistor $R_1$ and the parallel combination $R_p$: $R_{\text{ext}} = R_1 + R_p = 4\,\Omega + 9.375\,\Omega = 13.375\,\Omega$. For 17.3.3, the emf of the battery ($\varepsilon$) is calculated by adding the external potential difference to the internal potential drop. The external potential is $V_{\text{ext}} = I \times R_{\text{ext}} = 3\,A \times 13.375\,\Omega = 40.125\,V$. Therefore, $\varepsilon = V_{\text{ext}} + V_{\text{internal}} = 40.125\,V + 1.5\,V = 41.625\,V$. # Comparative Current Analysis and Resistor Removal Effects When evaluating the current through parallel branches, such as $R_2 = 25\,\Omega$ and $R_3 = 15\,\Omega$ in Question 17.4, the statement that the current through $R_3$ is larger than the current through $R_2$ is correct (YES). This is because current is inversely proportional to resistance in a parallel circuit where the potential difference across the branches is identical ($V = I \times R$). Since $R_3$ ($15\,\Omega$) has a lower resistance than $R_2$ ($25\,\Omega$), the lower resistance path will allow more current to flow. In Question 17.5, if the $4\,\Omega$ resistor ($R_1$) is removed from the circuit, the emf of the battery remains the same. This is because the emf is an intrinsic property of the chemical composition and design of the battery itself and is not dependent on the external load or the current flowing through the circuit. # Modeling Unknown Battery Parameters in Question 18 Question 18 details a circuit where the battery has both an unknown emf and an unknown internal resistance ($r$). Measurements show that with switch S open, the voltmeter reads $2.8\,V$, which represents the emf ($\varepsilon = 2.8\,V$). When switch S is closed, the voltmeter reads $2.63\,V$, representing the terminal potential difference ($V_{\text{ext}}$). The external circuit comprises a parallel pair of resistors, $4\,\Omega$ and $7\,\Omega$, connected in series with a $3\,\Omega$ resistor. To find the equivalent external resistance (18.2), we calculate the parallel branch: $R_p = \frac{4 \times 7}{4 + 7} = \frac{28}{11} \approx 2.55\,\Omega$. The total resistance is then $R_{\text{ext}} = 3\,\Omega + 2.55\,\Omega = 5.55\,\Omega$. To find the internal resistance (18.3.1), we first find the total current: $I = \frac{V_{\text{ext}}}{R_{\text{ext}}} = \frac{2.63\,V}{5.55\,\Omega} \approx 0.474\,A$. The lost volts are $V_{\text{lost}} = \varepsilon - V_{\text{ext}} = 2.8\,V - 2.63\,V = 0.17\,V$. Using the relationship $V_{\text{lost}} = I \times r$, we calculate $r = \frac{0.17\,V}{0.474\,A} \approx 0.359\,\Omega$. # Power and Potential in Open Circuits Question 19 describes a circuit with a battery of $\varepsilon = 12\,V$ and a configuration involving resistors of $6\,\Omega$, $2.4\,\Omega$, and $1.6\,\Omega$. A power rating of $5.76\,W$ is associated with the $6\,\Omega$ resistor. In 19.1.1, if switch S is open, voltmeter $V_1$ (which is connected across the battery) will read the full emf of the battery, which is $12\,V$. In 19.1.2, since the switch is open and no current flows through the circuit ($I = 0\,A$), the potential difference across any external resistor or voltmeter located after the break (like $V_2$) will be $0\,V$.