Question

A battery of emf 100 V and a resistor of resistance 10 kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100 Ω, the current through it is constant up to two significant digits.
Find its value. This is the basic principle of a constant-current source.

Answer 1

Given:
Emf of the battery, E = 100 volt
Resistance in series with battery, R' = 10 kΩ = 10000 Ω
External resistance, R = (1-100) Ω
When no external resistor is present (R = 0), current through the circuit,
$i=\frac{\mathrm{E}}{\mathrm{R}\mathit{\text{'}}}=\frac{100}{1000}=0.01\mathrm{Amp}.$
When R = 1 Ω,
$$\begin{gathered} i=\frac{100}{10000+1}=\frac{100}{10001} \\ =0.009999\mathrm{A} \end{gathered}$$
When R = 100 Ω,
$$\begin{gathered} i=\frac{100}{10000+100} \\ =\frac{100}{10100}=0.009900\mathrm{A} \end{gathered}$$
We can see that up to R = 100 Ω, the current does not change up to two significant digits.