Question

The internal resistance of a primary cell is 4 Ω. It generates a current of 0.2 A in an external resistance of 21Ω. The rate of chemical energy consumed in providing the current is
$$\begin{array}{rl}& \text{A}\text{. }\text{ 1 J/s}\\ & \text{B}\text{. }\text{ 5 J/s}\\ & \text{C}\text{. }\text{ 0}\text{.42 J/s}\\ & \text{D}\text{. }\text{ 0}\text{.8 J/s}\end{array}$$

Answer 1

Hint: The rate of chemical energy consumed is nothing but power. Use the formula of power. Many forms of power are available using any formula which will give you relation between current and resistance, since values and current resistance are available. Use Kirchhoff’s voltage and current law.

Formula used: $\text{power=(current)}\times \text{(current)}\times \text{(resistance)}$

Complete step by step answer:
First, understand what the question wants to convey.
Consider a circuit having cells and resistance connected to each other by wires. The cell is nothing but a device. Devices have their internal resistance. So the cell has resistance 4ohm. 0.2 is the current passing through the circuit. While external resistance that is the resistance which we have applied is 21 ohm.

Aim: The rate of chemical energy consumed in providing the current which is nothing but power.
We know that,
$\text{voltage=current }\times \text{ resistance}$
Mathematically is given by,
$V=I\times R--------\left(1\right)$
Use the formula of power I.e. The rate of chemical energy consumed in providing the current,
So by conservation of energy.
Heat energy produced per sec= Power

$\text{power=(current)}\times \text{(current)}\times \text{(resistance)}$
$\begin{array}{rl}& \text{power(P)=}{\text{I}}^{\text{2}}\text{(R+r)}\\ & \text{power=0}\text{.2 }\times \text{ 0}\text{.2 }\times \text{ (21+4)}\\ & \text{power=1J}{\text{s}}^{\text{-1}}\end{array}$

(Where r is the internal resistance and R is the external resistance)
Hence the rate of chemical energy consumed in providing the current is $\text{1J}{\text{s}}^{\text{-1}}$.
Hence, the correct answer is option A .

Note:
You can get the idea of what to find out in question by looking at the options. Internal resistance is the resistance which the device itself has while manufacturing. External resistance is what we apply to the circuit.