In a Wheatstone bridge, $P = 90\,\,{\rm{\Omega }},\,\,Q = 110\,\,{\rm{\Omega }},\,\,R = 40\,\,{\rm{\Omega }}\,\,{\rm{and}}\,\,S = 60\,\,{\rm{\Omega }}$ and a cell of ${\rm{4}}\,\,{\rm{V}}$ emf. Then the potential difference between the diagonal along which a galvanometer is connected is

NEET UG PHYSICS MCQ: CURRENT ELECTRICITY

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In a Wheatstone bridge,

P = 90\,\,\mathrm{\Omega },\,\,Q = 110\,\,\mathrm{\Omega },\,\,R = 40\,\,\mathrm{\Omega }\,\,\mathrm{and}\,\,S = 60\,\,\mathrm{\Omega }

and a cell of

\mathrm{4}\,\,\mathrm{V}

emf. Then the potential difference between the diagonal along which a galvanometer is connected is

View answer and explanation

Correct option: A

- 0.2\,\,\mathrm{V}

Option A (-0.2 V) is correct because it is the potential difference calculated using the Wheatstone bridge formula and given resistor values.

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