Question

A battery with negligible internal resistance is connected to a resistor. The power produced in the battery and power dissipated in the resistor are both P₁. Another resistor of same kind is added, so circuit consists of a battery and two resistors in series. (a) in terms of P₁ how much power is dissipated in the first resistor ? . (a) in terms of P₁ how much power is produced in the battery ? (c ) The circuit is rearranged so that the two resistors are in parallel rather than in series. In terms of P₁, now how much power is produced in the battery?

Short answer

The power dissipated across first resistor is $\frac{P_1}{2}$.

Step-by-step solution

Identification of given data

The power produced in battery and dissipated in resistor is $P_1$.

The number of resistors in series or parallel is $n=2$.

Conceptual Explanation

The power produced in the battery is the work done for the movement of charged particles across the terminals of the battery per unit time.

Determination of power dissipated in first resistor

The equivalent resistance of both resistors is given as:

$R_e=nR$

Here, $R$is the resistance of each resistor.

Substitute all the values in the above equation.

$R_e=2R$

The power produced in the battery is given as:

$P_1=I^2{R}_e$

Here, $I$ is the current of battery.

Substitute all the values in the above equation.

$$\begin{gathered} P_1=I^2\left(2R\right) \\ {P}_1=2\left(I^{2}R\right)......\left(1\right) \end{gathered}$$

The power produced across first resistor is given as:

$P=I^{2}R......\left(2\right)$

Divide the equation (1) by equation (2).

$\begin{array}{rcl}\frac{P_1}{P}& =& \frac{2\left(I^{2}R\right)}{\left(I^{2}R\right)}\\ P& =& \frac{P_1}{2}\\ & & \end{array}$

Therefore, the power dissipate in first resistor is half of the power produced in battery.