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Science · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 2 · Q39
(A) V-I graph for a nichrome wire is shown in the diagram. (i) Prove that the graph follows Ohm's law and calculate the resistance of nichrome wire…
V-I graph for a nichrome wire is shown in the diagram.
(i)
Prove that the graph follows Ohm's law and calculate the resistance of nichrome wire from the given V-I graph.
(ii)(I)
Calculate the equivalent resistance of the combination of resistors connected in the given circuit.
(II)
Find the value of the total current drawn from the battery.
(i)
Define the SI unit of power. Express the electric power of an electric appliance in terms of potential difference and current flowing through it.
(ii)
In a house, $\displaystyle 3$ bulbs of $\displaystyle 100$ W each, are lit for $\displaystyle 5$ hours daily and an electric heater of $\displaystyle 1.0$ kW is used for half an hour daily. Calculate the total energy consumed in a month of $\displaystyle 30$ days and its cost at the rate of ₹ $\displaystyle 3.60$ per kWh.
(iii)
Convert the commercial unit of electric energy into Joule (J).
Marking-scheme solution
(A) (i)
Since V-I graph is a straight line passing through the origin / V \(\displaystyle \propto \mathrm{I}\), it follows Ohm's Law.
R = Slope of V-I graph
\[R=\frac{0.8-0.4}{0.2-0.1}=\frac{0.4}{0.1}=4 \Omega
\]
(ii)(I)
\(\displaystyle 3 \Omega\) and \(\displaystyle 7 \Omega\) resistors are in series, \(\displaystyle \mathrm{R}_{\mathrm{s}}=3+7=10 \Omega \mathrm{R}_{\mathrm{s}}\) is in parallel combination with \(\displaystyle 10 \Omega\)
\[\begin{aligned}
& \frac{1}{R p}=\frac{1}{10}+\frac{1}{10} \\
& \operatorname{Rp}=5 \Omega
\end{aligned}
\]
Other two \(\displaystyle 5 \Omega\) resistors are in series with Rp.
\[\text { Net } \mathrm{R}=5+5+5=15 \Omega
\]
(II)
Total current, \(\displaystyle \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\)
\[\begin{aligned}
& I=\frac{5}{15} \\
& I=\frac{1}{3} A
\end{aligned}
\]
(B)
Power consumed by a device that carries $\displaystyle 1$ A of current when operated at a potential difference of $\displaystyle 1$ V / If one joule energy is consumed in one second then power of instrument is said to be $\displaystyle 1$ watt / \(\displaystyle 1 \mathrm{~W}=1\) volt \(\displaystyle \times 1\) ampere
P = VI
(ii)
\[\mathrm{E}=\mathrm{P} \times \mathrm{t}
\]
Energy \(\displaystyle (3\) bulbs \(\displaystyle )=3 \times 100 \times 5\)
\[\text { = } 1500
\]
\(\displaystyle =1.5 \mathrm{kWh}\)
Energy \(\displaystyle (\) electric heater \(\displaystyle )=1.0 \times 0.5\)
\[=0.5 \mathrm{kWh}
\]
Total energy consumed ($\displaystyle 1$ day) \(\displaystyle =1.5+0.5=2 \mathrm{kWh}\)
Total energy consumed ($\displaystyle 30$ days) \(\displaystyle =30 \times 2\)
\(\displaystyle =60 \mathrm{kWh}\)
\(\displaystyle =60\) units
Total cost \(\displaystyle =\) Units \(\displaystyle \times\) Rate
\(\displaystyle =60 \times 3.60\)
\(\displaystyle =₹ 216\)
(ii)
\(\displaystyle 1 \mathrm{kWh}=1000\) watt \(\displaystyle \times 3600\) second
\(\displaystyle =3.6 \times 10^{6}\) watt second
\(\displaystyle =3.6 \times 10^{6}\) joule (J)
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