CBSE 2026 · Region 4 · Set 3 · Q27 · 3 marks
(a)'Current is a scalar quantity, although we represent current with an arrow.' Explain.(b)Derive the balance condition of a Wheatstone Bridge.
(a)
'Current is a scalar quantity, although we represent current with an arrow.' Explain.
(b)
Derive the balance condition of a Wheatstone Bridge.
Marking-scheme solution
(a)
Current is a scalar quantity because it does not follow the law of vector addition. The current with an arrow is only to signify the direction of conventional current.
(b)
In the balanced condition, the current $\displaystyle I_{g}$ through G is 0. Then $\displaystyle \mathrm{I}_{1}=\mathrm{I}_{3}$ and $\displaystyle \mathrm{I}_{2}=\mathrm{I}_{4}$
For loop ADBA: $\displaystyle -\mathrm{I}_{1} \mathrm{R}_{1}+0+\mathrm{I}_{2} \mathrm{R}_{2}=0$, or $\displaystyle \dfrac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\dfrac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\quad \ldots(1)$
For loop CBDC, using $\displaystyle \mathrm{I}_{3}=\mathrm{I}_{1}$ and $\displaystyle \mathrm{I}_{4}=\mathrm{I}_{2}$: $\displaystyle \mathrm{I}_{2} \mathrm{R}_{4}+0-\mathrm{I}_{1} \mathrm{R}_{3}=0$, or $\displaystyle \dfrac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\dfrac{\mathrm{R}_{4}}{\mathrm{R}_{3}}\quad \ldots(2)$
From ($\displaystyle 1$) and ($\displaystyle 2$): $\displaystyle \dfrac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\dfrac{\mathrm{R}_{4}}{\mathrm{R}_{3}}$
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