CBSE 2026 · Region 2 · Set 2 · Q28 · 3 marks
(a)Explain the statement : "Current is a scalar although we represent current with an arrow".(b)Use Kirchhoff's rules to find the current through $\displaystyle 3 \Omega$ resistor in the circuit shown in the figure :
(a)
Explain the statement : "Current is a scalar although we represent current with an arrow".
(b)
Use Kirchhoff's rules to find the current through $\displaystyle 3 \Omega$ resistor in the circuit shown in the figure :
Marking-scheme solution
(a)
Electric current is a scalar quantity because it does not obey the law of vector addition. The current with an arrow is only to signify the direction of conventional current.
Alternatively: Current does not follow the law of vector addition. The current through an area of cross-section is given by the scalar product of two vectors, $\displaystyle \mathrm{I}=\overrightarrow{\mathrm{j}} \cdot \Delta \overrightarrow{\mathrm{S}}$
(b)
In loop ABEFA: $\displaystyle 3+3 \mathrm{I}_{1}-4\left(\mathrm{I}-\mathrm{I}_{1}\right)=0$
$\displaystyle 4 \mathrm{I}-7 \mathrm{I}_{1}=3 \quad \ldots(1)$
In loop BCDEB: $\displaystyle 5-2 \mathrm{I}-3 \mathrm{I}_{1}=0$
$\displaystyle 2 \mathrm{I}+3 \mathrm{I}_{1}=5 \quad \ldots(2)$
Solving equations ($\displaystyle 1$) and ($\displaystyle 2$), current through the $\displaystyle 3\ \Omega$ resistor is $\displaystyle \mathrm{I}_{1}=\dfrac{7}{13} \mathrm{~A}$
Current ElectricityKirchhoff’s RulesApplyshort_answermedium
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