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

Current (i) = 0.000999 Amp

Explanation:

From the question, given data as

Emf of the battery (E) = 100 V

Resistance (R) = 10 kΩ ; = 10000Ω

$\mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{100}{100000}=0.001 \mathrm{Amp}$

Step 1:

⇒When R = 1

$\mathrm{i}=\frac{100}{1000+1}=\frac{100}{1001}=0.09 \mathrm{A}$

Step 2:

⇒When R= 10

$\mathrm{i}=\frac{100}{10000+1}=\frac{100}{10001}=0.009 \mathrm{A}$

Step 3:

⇒ When R= 100

$\mathrm{i}=\frac{100}{100000+1}=\frac{100}{100001}=0.0009 \mathrm{A}$

Hence, we prove that up to resistance (R) = 100, current does not exceed up to 2 significant digits. Hence proved.