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

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

Step-by-step explanation:

Answer 2

The initial equivalent resistance of the circuit is

`(1)/(R_(eq))= (1)/(R_1)+ (1)/(R_2)= (1)/(8 \Omega)+ (1)/(8 \Omega) = (1)/(4 \Omega)`
which means

`R_(eq)= 4 \Omega`
Therefore the initial current in the circuit is

`I= (V)/(R)= (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

`(1)/(R_(eq)) = (1)/(8 \Omega) + (1)/(8 \Omega)+ (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= (V)/(R_(eq))= (9 V)/(3 \Omega)=3 A`
which means that the current in the circuit has increased.