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Science · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 2 · Q39
(i)Consider the following electric circuit :
Calculate the values of the following :(I)Total resistance of the circuit.(II)The total electric current drawn from the source.(III)Potential difference across $\displaystyle 3 \Omega$ resistor.(ii)Two bulbs, rated as $\displaystyle 100 \mathrm{~W} ; 220 \mathrm{~V}$ and $\displaystyle 60 \mathrm{~W} ; 220 \mathrm{~V}$ are connected in parallel to an electric main supply of $\displaystyle 220$ V. Calculate the electric current drawn from the mains.(i)State Ohm's law and draw V-I graph for a conductor which follows Ohm's law. Show that the slope of V-I graph gives resistance of conductor.(ii)Derive an expression for the equivalent resistance of a series combination of three resistors having resistances $\displaystyle \mathrm{R}_{1}, \mathrm{R}_{2}$ and $\displaystyle \mathrm{R}_{3}$.
(i)
Consider the following electric circuit :
Calculate the values of the following :
(I)
Total resistance of the circuit.
(II)
The total electric current drawn from the source.
(III)
Potential difference across $\displaystyle 3 \Omega$ resistor.
(ii)
Two bulbs, rated as $\displaystyle 100 \mathrm{~W} ; 220 \mathrm{~V}$ and $\displaystyle 60 \mathrm{~W} ; 220 \mathrm{~V}$ are connected in parallel to an electric main supply of $\displaystyle 220$ V. Calculate the electric current drawn from the mains.
(i)
State Ohm's law and draw V-I graph for a conductor which follows Ohm's law. Show that the slope of V-I graph gives resistance of conductor.
(ii)
Derive an expression for the equivalent resistance of a series combination of three resistors having resistances $\displaystyle \mathrm{R}_{1}, \mathrm{R}_{2}$ and $\displaystyle \mathrm{R}_{3}$.
Marking-scheme solution
(a)
(i) (I)
\[\text { (I) } \quad \begin{aligned}
& \mathrm{R}_{\mathrm{S}}=2+2=4 \Omega \\
& \frac{1}{\mathrm{R}^{\prime}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4} \\
& \mathrm{R}^{\prime}=2 \Omega \\
& \mathrm{R}^{\prime \prime}=\mathrm{R} \prime+3 \Omega \\
& \mathrm{R}^{\prime \prime}=2 \Omega+3 \Omega \\
& \mathrm{R}^{\prime \prime}=5 \Omega
\end{aligned}
\]
Potential difference across \(\displaystyle \mathrm{R}_{1}, \mathrm{R}_{2}, \mathrm{R}_{3}\) is
\[\mathrm{V}_{1}=\mathrm{IR}_{1}, \mathrm{~V}_{2}=\mathrm{IR}_{2}, \mathrm{~V}_{3}=\mathrm{IR}_{3}
\]
If, \(\displaystyle \mathrm{R}_{\mathrm{S}}\) is equivalent resistance in series combination and I is the current through the circuit then
\[V=I R_{S}
\]
\[1 \text { ne total potential dirrerence, }
\]
\[V=V_{1}+V_{2}+V_{3}
\]
\[\mathrm{IR}_{\mathrm{S}}=\mathrm{IR}_{1}+\mathrm{IR}_{2}+\mathrm{IR}_{3}
\]
\[\mathrm{R}_{\mathrm{S}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}
\]
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CBSE Class 10 Science past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.