A battery of emf 100 V and a resistor of resistance 10 k - 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 - the current through it is constant up to two significant digits- Find its value- This is the basic principle of a constant-currentsource-

E = 100 volt, R' = 100 k Ω = 100000 Ω R = 1 upto 100 When no other register is added or r = 0 When r = 1 i = E/R = 100/10000 = 0.000999 A up to r = 100, the cur ...

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Byju's Answer

Standard XII

Physics

Resistors in Series and Parallel

A battery of ...

Question

A battery of emf 100 V and a resistor of resistance 10 k $Ω$ 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.

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Solution

E = 100 volt,

R' = 100

$k Ω = 100000 Ω$

R = 1 upto 100 When no other register is added or r = 0

When r = 1

i = $\frac{E}{R} = \frac{100}{10000}$

= 0.000999 A

up to r = 100, the current does not up to two significant digits. Thus it is proved.

When R = 10.

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

= 0.000999 A.

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E = 100 volt, R' = 100 kΩ=100000Ω R = 1 upto 100 When no other register is added or r = 0 When r = 1 i = ER=10010000 = 0.000999 A up to r = 100, the current does not up to two significant digits. Thus it is proved. When R = 10. i = 100100000+1=100100001 = 0.000999 A.