Question

a parallel circuit has two 8.0 ohm resistors and a power source of 9.0 volts. Of 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

Current = 9 / 28.5
current = .315...

Answer 2

Answer : The new current has increased and the value of current will be 3.0 A.

Explanation :

Equivalent resistance : It represents the total effect of all resistors in the circuit.

In series circuit, we add all the resistances of each component.

$R=R_1+R_2+R_3....$

In parallel circuit, the reciprocal of the total resistance is equal to the sum of the reciprocals of the resistances of each component.

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

First we have to calculate the current when a parallel circuit has two 8.0 ohm resistors and a power source of 9.0 volts.

As we are given:

$R_1=8\Omega$

$R_2=8\Omega$

Now we have to calculate the equivalent resistance for a parallel circuit.

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{1}{8}+\frac{1}{8}$

$R=4\Omega$

Now we have to calculate the initial current in the circuit.

$I=\frac{V}{R}=\frac{9.0V}{4\Omega}=2.25A$

When the new resistor of 12.5 ohm is added to the circuit in parallel, the new equivalent resistance of the circuit is:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R}=\frac{1}{8}+\frac{1}{8}+\frac{1}{12.5}$

$R=3.0\Omega$

Now we have to calculate the new current in the circuit when the new resistor of 12.5 ohm is added.

$I=\frac{V}{R}=\frac{9.0V}{3.0\Omega}=3.0A$

This means that the equivalent resistance of the circuit has decreased, and the new current has increased.

Hence, the new current has increased and the value of current will be 3.0 A.