# How do you solve for the voltages and currents in this circuit diagram?

## Jun 26, 2016

#### Explanation:

As ${R}_{1}$ and ${R}_{2}$ are in parallel, the resultant resistance $R$ will be given by $\frac{1}{{R}_{1}} + \frac{1}{{R}_{2}}$.

Hence $\frac{1}{R} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ and hence $R = 2$ $\Omega$.

As the voltage of battery source is $120 V$, assuming $0$ internal resistance, current passing through circuit or ${R}_{3}$ is ${I}_{0} = \frac{120}{15 + 2} = \frac{120}{17} = 7 \frac{1}{17}$ ampere.

and hence voltage ${V}_{3}$ across it will be $\frac{120}{17} \times 15 = \frac{1800}{17} = 105 \frac{15}{17} V$

and that across other two resistances i.e. ${V}_{1} = {V}_{2}$ will be $120 - \frac{1800}{17} = \frac{240}{17} V$ each.

Hence current $I - 1$ passing through ${R}_{1}$ is $\frac{240}{17} \times \frac{1}{3} = \frac{80}{17} = 4 \frac{12}{17}$ ampere.

and current passing ${I}_{2}$ through ${R}_{2}$ is $\frac{240}{17} \times \frac{1}{2} = \frac{120}{17} = 7 \frac{1}{17}$ ampere.