DC Circuits Practice

The simulation below allows students to practise calculating potential differences and currents of a slightly complex circuit, involving three different modes that can be toggled by clicking on the switch.

Link: https://www.geogebra.org/m/jkckp9pr

Mode 1: Two Resistors in Series

When resistors $R_1$ and $R_2$ are connected in series, the total resistance is simply the sum of the individual resistances:

$$R_{\text{total}}=R_1+R_2$$

The current $I$ through the circuit is given by Ohm’s Law:

$$I=\frac{V_{\text{total}}}{R_{\text{total}}}=\frac{V_{\text{total}}}{R_1+R_2}$$

where $V_{\text{total}}$ is the total potential difference supplied by the source.

The potential difference across each resistor can be calculated using:

$$V_1=I\cdot R_1,V_2=I\cdot R_2$$

Mode 2: $R_1$ and $R_3$ in Parallel, $R_2$ in Series

In this mode, resistors $R_1$ and $R_3$ are in parallel, and $R_2$ is in series with the combination. First, calculate the equivalent resistance of the parallel combination:

$$\frac{1}{R_{\text{parallel}}}=\frac{1}{R_1}+\frac{1}{R_3}$$

Thus, the total resistance is:

$$R_{\text{total}}=R_{\text{parallel}}+R_2$$

The current through the circuit is:

$$I=\frac{V_{\text{total}}}{R_{\text{total}}}$$

The potential difference across $R_2$ is:

$$V_2=I\cdot R_2$$

Since $R_1$ and $R_3$ are in parallel, they share the same potential difference:

$$V_1=V_3=V_{\text{total}}–V_2$$

The current through each parallel resistor can be found using Ohm’s Law:

$$I_1=\frac{V_1}{R_1},I_3=\frac{V_3}{R_3}$$

Mode 3: $R_1$ and $R_2$ in Series, $R_3$ in Parallel

Here, resistors $R_1$ and $R_2$ are connected in series, and the combination is in parallel with $R_3$. First, calculate the resistance of the series combination:

$$R_{\text{series}}=R_1+R_2$$

Then, find the total resistance of the parallel combination:

$$\frac{1}{R_{\text{total}}}=\frac{1}{R_{\text{series}}}+\frac{1}{R_3}$$

The total current is:

$$I=\frac{V_{\text{total}}}{R_{\text{total}}}$$

The voltage across the parallel combination is the same for both branches:

$$V_1+V_2=V_3=V_{\text{total}}$$

The current through $R_3$ is:

$$I_3=\frac{V_3}{R_3}$$

The current through $R_1$ and $R_2$, which are in series, is the same:

$$I_{\text{series}}=\frac{V_{\text{total}}}{R_1+R_2}$$

The voltage across each series resistor is:

$$V_1=I_{\text{series}}\cdot R_1,V_2=I_{\text{series}}\cdot R_2$$