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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.

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प्रश्न

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.

योग
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उत्तर

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,

\[i = \frac{E}{R'} = \frac{100}{1000} = 0 . 01 \text{ Amp.}\]

When R = 1 Ω,

\[i   =   \frac{100}{10000 + 1} = \frac{100}{10001}\]

\[     = 0 . 009999  A\]

When R = 100 Ω,

\[i = \frac{100}{10000 + 100}\]

\[   = \frac{100}{10100} = 0 . 009900  A\]

We can see that up to R = 100 Ω, the current does not change up to two significant digits.

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अध्याय 32: Electric Current in Conductors - Exercises [पृष्ठ १९९]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 32 Electric Current in Conductors
Exercises | Q 22 | पृष्ठ १९९

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