Question

trying to get exempt from the final.

3. A 9.0 V battery is hooked up with three resistors (R1, R2, R3) in parallel with resistances of 3.0 Ω, 7.0 Ω, and 6.0 Ω, respectively.
a.) Draw a labeled circuit diagram for the circuit described.
b.) Calculate the equivalent resistance.
c.) Calculate the current passing through each resistor in the circuit.
Answer:

Answer 1

Can you draw the circuit?

Answer 2

No dont know how.

Answer 3

Quick reminder: Parallel means they can be drawn side-by-side. Series means they are one after the other. For two resistors connected to a battery:

Parallel:

Series:

Using that, you should be able to draw the circuit.

Since they are in parallel, we find the equivalent resistance by doing:
\[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}+\frac{1}{R_3}\]Just solve for Req. (Remember to solve for Req, NOT for 1/Req. This is a common mistake)

Since the resistors are all in parallel, they have the same voltage drop across them. This voltage drop is equivalent to the voltage from the battery, since there is nothing else changing that voltage in the circuit. To find the current across each resistor, we can just use V=IR for each individual resistor.

That is:
\[V=I \times R_1\]will give the current going across resistor 1.
You can check your answers by using V=IR with the equivalent resistance. If the current going across the equivalent resistor is equal to the sum of the currents across the individual resistors, then you are likely correct in your work.