Chapter 23.2, Problem 25PP

To determine

The total current drawn from the battery by the circuit

Answer to Problem 25PP

The current drawn from the battery is $I=0.54\text{ A}$

Explanation of Solution

Given:

A circuit consisting of two parallel branches of resistors, connected in series with another resistor across a 12-V battery. The series resistor uses a power of $P_1=2\text{ W}$ , the first resistor in the parallel branch uses a power of $P_2=3.0\text{ W}$ and the third resistor uses a power of $P_3=1.5\text{ W}$ .

Formula used:

The total power supplied by the source is equal to the total power dissipated in the resistors. The total power supplied by a battery of voltage V and supplying a total current of I is given by,

  $P_s=VI$ .... (1)

Now, the total power dissipated is given by,

  $P_d=P_1+P_2+P_3$

Thus, it implies that,

  $P_s=P_d$ .... (2)

Calculation:

Consider the circuit shown in Figure 1. Substituting the given power values, the power dissipated is given by,

  $\begin{array}{c}{P}_d=2+3+1.5\\ =6.5\text{}\text{ W}\end{array}$

Now, using the relations (1) and (2) and substituting the supply voltage,

  $\begin{array}{c}6.5=12I\\ I=\frac{6.5}{12}\\ =0.54\text{ A}\end{array}$

Conclusion:

The total current drawn from the battery is 0.54 A.