GSO001 Problem 6

This problem is solved by computing the combined resistance, distributing the current by Ohm's law, and then applying the Joule heat formula.

First, find the combined resistance $R_{23}$ of the parallel section ( $R_2$ and $R_3$ ):

\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} \implies R_{23} = 4 \Omega

The total resistance $R_{total}$ of the circuit is:

R_{total} = R_1 + R_{23} = 4 + 4 = 8 \Omega

The total current $I$ from the source by Ohm's law:

I = \frac{V}{R_{total}} = \frac{48}{8} = 6 \text{ A}

The voltage $V_{p}$ across the parallel section:

V_{p} = I \times R_{23} = 6 \times 4 = 24 \text{ V}

(Alternatively: $V_{p} = V - I \cdot R_1 = 48 - 6 \times 4 = 24 \text{ V}$ )

Since it is a parallel circuit, the voltage across $R_3$ is $V_{p}$ . The power $P_3$ consumed in $R_3$ :

P_3 = \frac{V_{p}^2}{R_3} = \frac{24^2}{6} = \frac{576}{6} = 96 \text{ W}

The Joule heat $Q_3$ generated during time $t$ :

Q_3 = P_3 \times t = 96 \times 15 = 1440 \text{ J}

Answer: 1440

Energy Consumed in a Combined Resistor Circuit