Electricity and Magnetism - Electric circuits

A battery with an EMF of $12.0 V$ and an internal resistance of $0.5 \Omega$ is connected to an external resistor of $5.5 \Omega$. Calculate the terminal potential difference across the battery.

First, find the total resistance: $R_{total} = R + r = 5.5 \Omega + 0.5 \Omega = 6.0 \Omega$. Next, calculate the circuit current: $I = \frac{\epsilon}{R_{total}} = \frac{12.0 V}{6.0 \Omega} = 2.0 A$. Finally, calculate the terminal voltage: $V = \epsilon - Ir = 12.0 V - (2.0 A \times 0.5 \Omega) = 11.0 V$.

The terminal potential difference is always less than the EMF when a current flows because of the voltage drop across the internal resistance ($Ir$).

Calculate the total resistance of three resistors ($2 \Omega$, $4 \Omega$, and $4 \Omega$) connected in parallel.

Using the parallel resistance formula: $\frac{1}{R_{total}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{4}$. This simplifies to $\frac{1}{R_{total}} = \frac{2}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4} = 1 \Omega^{-1}$. Therefore, $R_{total} = 1 \Omega$.

In a parallel circuit, the total resistance is always less than the smallest individual resistance.

A $2.0 kW$ electric kettle is connected to a $250 V$ mains supply. Calculate the current flowing through the element and its resistance.

Using $P = VI$, current $I = \frac{P}{V} = \frac{2000 W}{250 V} = 8.0 A$. Using $V = IR$, resistance $R = \frac{V}{I} = \frac{250 V}{8.0 A} = 31.25 \Omega$.

Power must be converted to Watts ($1 kW = 1000 W$) before using the standard formulae.

In the following series-parallel circuit, find the total current $I$ supplied by the $12 V$ battery. The resistors are $R_1 = 4 \Omega$, $R_2 = 6 \Omega$, and $R_3 = 3 \Omega$.

1. Find the equivalent resistance of the parallel branch ($R_2$ and $R_3$): $\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \Omega^{-1}$ $R_p = 2 \Omega$
2. Calculate total resistance ($R_{total}$): $R_{total} = R_1 + R_p = 4 + 2 = 6 \Omega$
3. Use Ohm's Law to find total current: $I = \frac{V}{R_{total}} = \frac{12}{6} = 2 A$

We first simplify the parallel combination of $R_2$ and $R_3$ into a single equivalent resistor, then add it to $R_1$ because they are in series. Finally, we use the battery voltage and total resistance to find the source current.

A potential divider circuit is used to provide a variable output voltage. The input voltage is $V_{in} = 9.0 V$, and the circuit consists of a fixed resistor $R_1 = 1200 \Omega$ and a thermistor $R_2$. At a certain temperature, the resistance of the thermistor is $R_2 = 600 \Omega$. Calculate the output voltage $V_{out}$ across the thermistor and determine the current flowing through the circuit.

$V_{out} = V_{in} \times \left( \frac{R_2}{R_1 + R_2} \right)$ $V_{out} = 9.0 \times \left( \frac{600}{1200 + 600} \right)$ $V_{out} = 9.0 \times \left( \frac{600}{1800} \right)$ $V_{out} = 9.0 \times \frac{1}{3} = 3.0 V$

For the current $I$: $I = \frac{V_{in}}{R_{total}}$ $I = \frac{9.0}{1200 + 600}$ $I = \frac{9.0}{1800} = 0.005 A = 5.0 mA$

In a potential divider circuit, the input voltage is shared between resistors in series in proportion to their resistance. The output voltage across the second resistor is found using the potential divider formula. The total current is found by dividing the supply voltage by the sum of all resistances in the series loop.