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Science · 2024 · 5 marks
CBSE 2024 · Region 5 · Set 3 · Q36
(a) (i) Which type of circuits - series or parallel, should be used when you have to operate different electrical gadgets in your house? List two…
(i)
Which type of circuits - series or parallel, should be used when you have to operate different electrical gadgets in your house ? List two reasons for your answer.
(ii)
Find the current flowing through the following electric circuit:
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given in the following table :
| I (Amperes) | $\displaystyle 0 \cdot 5$ | $\displaystyle 1 \cdot 0$ | $\displaystyle 2 \cdot 0$ | $\displaystyle 3 \cdot 0$ | $\displaystyle 3 \cdot 5$ |
| V (Volts) | $\displaystyle 1$•$\displaystyle 5$ | $\displaystyle 3 \cdot 0$ | $\displaystyle 6 \cdot 2$ | $\displaystyle 9 \cdot 3$ | $\displaystyle 10 \cdot 8$ |
(i)
Plot a graph between V and I.
(ii)
Calculate the resistance of the resistor with the help of the graph.
(iii)
What does the graph represent ?
(iv)
Why should this graph pass through the origin ?
Marking-scheme solution
(a) (i) Parallel Circuit
Each electrical appliance has its own switch due to which it can be turned ON and OFF separately.
If one electrical appliance stops working, others remain unaffected.
Each appliance has equal potential difference and draws current as per its requirement.
The total resistance in parallel circuit decreases.
(ii) Combined resistance of the series,
\[\mathrm{R}_{1}=6 \Omega+6 \Omega=12 \Omega
\]
Combined resistance of parallel grouping of \(\displaystyle 6 \Omega\) and \(\displaystyle \mathrm{R}_{1}=12 \Omega\), resistors is \(\displaystyle \mathrm{R}_{2}\), where
\[\frac{1}{\mathrm{R}_{2}}=\frac{1}{6}+\frac{1}{12}=\frac{9}{20}
\]
\[\mathrm{R}_{2}=4 \cdot 0 \Omega
\]
Total resistance of circuit \(\displaystyle =\mathrm{R}=3+4+3=10 \Omega\)
Current flowing \(\displaystyle =\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\)
\[\begin{aligned}
& =\frac{4.5 \mathrm{~V}}{10 \Omega} \\
& =0.45 \mathrm{~A}
\end{aligned}
\]
(ii) Resistance of resistor \(\displaystyle =\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{I}_{2}-\mathrm{I}_{1}}=\frac{8 \cdot 3-5 \cdot 2}{2 \cdot 5-1 \cdot 5}=3 \cdot 1 \Omega\)
(iii) The given resistor obeys Ohm's law./ Resistance remains constant.
(iv) Because when the value of \(\displaystyle \mathrm{V}=0\), the current \(\displaystyle \mathrm{I}=0\).
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CBSE Class 10 Science past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.