In a Wheatstone bridge, P = 90,Omega , Q = 110,Omega , R = 40,Omega and S = 60,Omega and a cell of 4,V,emf. Then

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

b $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{AB}}=\frac{4}{90+110} \times 90=\frac{9}{5}$ ............$(i)$ $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{AC}}=\frac{4}{40+60} \times 40=\frac{8}{5}$ .........$(ii)$ eq. $(i)$ $-$ eq. $(ii)$ $\Rightarrow \mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{B}}=\frac{9}{5}-\frac{8}{5}=\frac{1}{5}=0.2 \mathrm{\,V}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$0$
B$0.2$
C$0.5$
D$1$

Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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