Question

Q. A student is connected a dry cell of 1 point
emf 1.5V to an external resistance of
10. When he measured the potential
difference across the terminals of the
cell, voltmeter reading was 1.4V then
the internal resistance of voltmeter
will be: *
O
0.85 ohm
0
2.4 ohm
O
1.6 ohm
O
0.71 ohm​

Answer 1

Internal resistance of  Battery  = 0.71  Ω  if 1.5V connected with 10 Ω Resistance and voltmeter across that reads 1.4V

Explanation:

Correction in Question :

 internal resistance of  Battery to be found   not of voltmeter

Emf = 1.5 V

Voltmeter measure 1.4V across 10 Ω Resistance

Hence current through 10 Ω Resistance = 1.4/10 = 0.14 A

Emf -  Ir   = IR

r =  internal resistance of Battery

R = External Resistance

=> 1.5  - 0.14r  = 0.14(10)

=> 1.5 - 0.14r = 1.4

=> 0.14r  = 0.1

=> r = 0.1/0.14

=> r = 10/14

=> r = 5/7

=> r = 0.71  Ω

internal resistance of  Battery  = 0.71  Ω

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