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.

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

$I\propto\frac{1}{R}$

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

$\frac{1}{R_p}=\frac{1}{R1}+\frac{1}{R2}$

(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

$\begin{gathered} \frac{1}{R_d}=\frac{1}{R}+\frac{1}{R} \\ =\frac{2}{R} \\ R_d=\frac{R}{2} \end{gathered}$

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.