Question

Three 15.0-W resistors are connected in parallel across a 30.0-V battery. a) Find the current through each branch of the circuit. b) Find the equivalent resistance of the circuit.c) Find the current through the battery.

Answer 1

First, find the net resistance between the parallel resistors.

$\frac{1}{R_{\text{net}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Use the given magnitudes.

$\begin{gathered} \frac{1}{R_{\text{net}}}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{1+1+1}{15}=\frac{3}{15}=\frac{1}{5} \\ R_{\text{net}}=5\Omega \end{gathered}$

(b) The net resistance is 5 ohms.

Then, find the current using Ohm's Law.

$\begin{gathered} I=\frac{V}{R} \\ I=\frac{30V}{5\Omega} \\ I=6A \end{gathered}$

(c) The current through the battery is 6 A.

To find the current through each branch, we have to use Ohm's Law for each resistor.

$\begin{gathered} I=\frac{V}{R}=\frac{30V}{15\Omega} \\ I=2A \end{gathered}$

(a) The current passing through each branch of the circuit is 2 A.