14.An ammeter together with an unknown
resistance in series is connected across two
identical batteries each of emf 1.5V.Whenthe
batteries are connected in series ,the
galvanometer records a current of 1A and when
the batteries are in parallel ,the current is 0.6A.
What is the internal resistance of each battery?
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Explanation:
Given An ammeter together with an unknown resistance in series is connected across two identical batteries each of emf 1.5 V. When the
batteries are connected in series ,the galvanometer records a current of 1 A and when the batteries are in parallel ,the current is 0.6 A. What is the internal resistance of each battery?
- So assume R as combined resistance of galvanometer and unknown resistance and r will be internal resistance of each battery.
- So current i1 = 2 E / R + 2r
- So 1 = 2 x 1.5 / R + 2r
- So R + 2r = 3 ------------1
- Now when the batteries are connected in parallel we have
- Net resistance will be r/2
- Therefore i2 = E / R + r/2 (r net = r x r / r + r, so r net = r/2)
- = 2 E / 2 R + r
- Or 2 R + r = 2 E / i2
- 2R + r = 2 x 1.5 / 0.6
- 2 R + r = 5 -----------2
- From 1 and 2 we get
- R + 2r = 3
- 2R + r = 5
- Multiply eqn 1 by 2 we get
- 2R + 4r = 6
- 2R + r = 5
- So 3r = 1
- Or r = 1/3 ohm
- So internal resistance r = 1/3 ohm
Reference link will be
https://brainly.in/question/9756087
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