Question

Which of the following circuits has the greatest amount of current leaving the battery? Assume all batteries and light bulbs are identical.Group of answer choicesA series circuit with 2 light bulbsA series circuit with 3 light bulbsA simple circuit with one light bulbA parallel circuit with two light bulbs

Answer 1

Given:

(a) A series circuit with 2 light bulbs

(b) A series circuit with 3 light bulbs

(c) A simple circuit with one light bulb

(d) A parallel circuit with two light bulb

Also, light bulbs are identical.

To find the circuit having the greatest amount of current leaving the battery.

Step-by-step explanation:

The light bulb acts as a resistance in the circuit.

As light bulb are identical, let each bulb has R resistance.

The battery supplied is the same, so the voltage is the same.

According to Ohm's law,

`\begin{gathered} V\propto\text{ I} \\ V=IR \end{gathered}`

As the voltage is constant, so the current will be

`I\propto(1)/(R)`

We have to calculate the resistance of each circuit, in order to find the greatest current in the circuit.

The series resistance can be calculated by the formula

`R_s=R1+R2`

The parallel resistance can be calculated by the formula

`(1)/(R_p)=(1)/(R1)+(1)/(R2)`

(a) The resistance of the series circuit with 2 light bulbs will be

`\begin{gathered} R_a=R+R \\ =2R \end{gathered}`

(b) The resistance of the series circuit with 3 light bulbs will be

`\begin{gathered} R_b=R+R+R \\ =3R \end{gathered}`

(c) The simple circuit has resistance

`R_c=R`

(d) The resistance of the circuit with two resistances will be

`\begin{gathered} (1)/(R_d)=(1)/(R)+(1)/(R) \\ =(2)/(R) \\ R_d=(R)/(2) \end{gathered}`

As the resistance of the parallel circuit with two light bulbs is minimum, so the current will be maximum.

Thus, the circuit (d) parallel circuit with two light bulbs has a maximum current leaving the battery.