CBSE 2026 · Region 5 · Set 1 · Q31 · 5 marks
(i)Derive the condition for which a Wheatstone Bridge is balanced.(ii)Determine the current in $\displaystyle 3 \Omega$ branch of a Wheatstone Bridge in the circuit shown in the figure.
(i)Consider a cylindrical conductor of length $\displaystyle l$ and area of cross-section A. Current I is maintained in the conductor and electrons drift with velocity $\displaystyle \mathrm{v}_{\mathrm{d}}\left(\left|\overrightarrow{\mathrm{v}}_{\mathrm{d}}\right|=\frac{\mathrm{e}|\overrightarrow{\mathrm{E}}|}{\mathrm{m}} \tau\right)$, (where symbols have their usual meanings). Show that the conductivity $\displaystyle \sigma$ of the material of the conductor is given by $\displaystyle \sigma=\frac{\mathrm{ne}^{2}}{\mathrm{~m}} \tau$.(ii)The resistance of a metal wire at $\displaystyle 20^{\circ} \mathrm{C}$ is $\displaystyle 1.05 \Omega$ and at $\displaystyle 100^{\circ} \mathrm{C}$ is $\displaystyle 1.38 \Omega$. Determine the temperature coefficient of resistivity of this metal.
(i)
Derive the condition for which a Wheatstone Bridge is balanced.
(ii)
Determine the current in $\displaystyle 3 \Omega$ branch of a Wheatstone Bridge in the circuit shown in the figure.
(i)
Consider a cylindrical conductor of length $\displaystyle l$ and area of cross-section A. Current I is maintained in the conductor and electrons drift with velocity $\displaystyle \mathrm{v}_{\mathrm{d}}\left(\left|\overrightarrow{\mathrm{v}}_{\mathrm{d}}\right|=\frac{\mathrm{e}|\overrightarrow{\mathrm{E}}|}{\mathrm{m}} \tau\right)$, (where symbols have their usual meanings). Show that the conductivity $\displaystyle \sigma$ of the material of the conductor is given by $\displaystyle \sigma=\frac{\mathrm{ne}^{2}}{\mathrm{~m}} \tau$.
(ii)
The resistance of a metal wire at $\displaystyle 20^{\circ} \mathrm{C}$ is $\displaystyle 1.05 \Omega$ and at $\displaystyle 100^{\circ} \mathrm{C}$ is $\displaystyle 1.38 \Omega$. Determine the temperature coefficient of resistivity of this metal.
Marking-scheme solution
a.(i)
Applying Kirchoff's rule to closed loop ADBA. For balanced condition $\displaystyle \mathrm{Ig}=0$.
\[-\mathrm{I}_{1} \mathrm{R}_{1}+0+\mathrm{I}_{2} \mathrm{R}_{2}=0
\]
Applying Kirchoff's rule to closed loop CBDC using $\displaystyle \mathrm{I}_{3}=\mathrm{I}_{1}, \mathrm{I}_{4}=\mathrm{I}_{2}$
\[\mathrm{I}_{2} \mathrm{R}_{4}+0-\mathrm{I}_{1} \mathrm{R}_{3}=0 .
\]
From eq.(i).
\[\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}
\]
and from eq.(ii)
\[\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{4}}{\mathrm{R}_{3}}
\]
hence $\displaystyle \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{R}_{4}}{\mathrm{R}_{3}}$
Applying Kirchoff's rule to closed loop ADBA. For balanced condition $\displaystyle \mathrm{Ig}=0$.
\[-\mathrm{I}_{1} \mathrm{R}_{1}+0+\mathrm{I}_{2} \mathrm{R}_{2}=0
\]
Applying Kirchoff's rule to closed loop CBDC using $\displaystyle \mathrm{I}_{3}=\mathrm{I}_{1}, \mathrm{I}_{4}=\mathrm{I}_{2}$
\[\mathrm{I}_{2} \mathrm{R}_{4}+0-\mathrm{I}_{1} \mathrm{R}_{3}=0 .
\]
From eq.(i).
\[\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}
\]
and from eq.(ii)
\[\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{4}}{\mathrm{R}_{3}}
\]
hence $\displaystyle \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{R}_{4}}{\mathrm{R}_{3}}$Current ElectricityWheatstone BridgeApplylong_answerhard
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