Question

A parallel circuit has two 8.0-ohm resistors and a power source of 9.0 volts. If a 12.5-ohm resistor is added to the circuit in parallel, how will the current be affected and what value will it have?

Answer 1

The initial equivalent resistance of the circuit is
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}= \frac{1}{8 \Omega}+ \frac{1}{8 \Omega} = \frac{1}{4 \Omega}$
which means 
$R_{eq}= 4 \Omega$
Therefore the initial current in the circuit is
$I= \frac{V}{R}= \frac{9 V}{4 \Omega}=2.25 A$

When the new resistor of $12.5 \Omega$ is added to the circuit in parallel, the new equivalent resistance of the circuit is
$\frac{1}{R_{eq}} = \frac{1}{8 \Omega} + \frac{1}{8 \Omega}+ \frac{1}{12.5 \Omega}= 0.33 \Omega^{-1}$
from which we find
$R_{eq}=3 \Omega$
This means that the equivalent resistance of the circuit has decreased, and the new current is
$I= \frac{V}{R_{eq}}= \frac{9 V}{3 \Omega}=3 A$
which means that the current in the circuit has increased.

Answer 2

Answer:

1/R1 + 1/R2 + ... = 1/Re

So...

1/17.2 + 1/22.4 = 1/Re

0.1021 = 1/Re

Re = 9.792 Ohms

Now use the Voltage equation V = IR

6 = I * 9.792

I = 6/9.792 = 0.613 Amps.Pato 0.61

Explanation: