Wheat stone bridge is balanced. Find ratio of electric power dissipated in (P + Q) to (R + S)

Q: Wheat stone bridge is balanced. Find ratio of electric power dissipated in $(P + Q)$ to $(R + S)$

b $\frac{\text { Power in }(\mathrm{P}+\mathrm{Q})}{\text { Power in }(\mathrm{R}+\mathrm{S})}=\frac{\frac{\mathrm{V}^{2}}{\mathrm{P}+\mathrm{Q}}}{\frac{\mathrm{V}^{2}}{\mathrm{R}+\mathrm{S}}}=\frac{\mathrm{R}+\mathrm{S}}{\mathrm{P}+\mathrm{Q}}$ $=\frac{\mathrm{R}\left[1+\frac{\mathrm{S}}{\mathrm{R}}\right]}{\mathrm{P}\left[1+\frac{\mathrm{Q}}{\mathrm{P}}\right]}$ ........$(A)$ Also $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}} \Rightarrow \frac{\mathrm{S}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{P}}$ ...........$(B)$ from $(A)$ and $(B)$ $\frac{\text { Power in }(\mathrm{P}+\mathrm{Q})}{\text { Power in }(\mathrm{R}+\mathrm{J})}=\frac{\mathrm{R}}{\mathrm{P}}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Wheat stone bridge is balanced. Find ratio of electric power dissipated in $(P + Q)$ to $(R + S)$

A$1 : 1$
B$R : P$
C$P^2 : Q^2$
D$P^2 : R^2$

Medium

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

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