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}}}=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)`

Use the given magnitudes.

`\begin{gathered} \frac{1}{R_{\text{net}}}=(1)/(15)+(1)/(15)+(1)/(15)=(1+1+1)/(15)=(3)/(15)=(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=(V)/(R) \\ I=(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=(V)/(R)=(30V)/(15\Omega) \\ I=2A \end{gathered}`

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