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Science · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 1 · Q39
(i)The given electric circuit is a part of an electrical device. Use the information given in the electric circuit diagram to calculate :(I)Potential difference across the ends of resistor $\displaystyle \mathrm{R}_{2}$.(II)Value of resistor $\displaystyle \mathrm{R}_{2}$.(III)Value of resistor $\displaystyle \mathrm{R}_{1}$.(ii)Write the factors on which resistance of a conductor depends and derive the formula for resistance of a given conductor.(B)How much electric current will an electric iron draw from $\displaystyle 220$ V source if the resistance of its heating element when hot, is $\displaystyle 55 \Omega$ ? Calculate the power consumed by the electric iron when it is operated at $\displaystyle 220$ V.(ii)In a house, $\displaystyle 3$ bulbs of $\displaystyle 100$ watt 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)With reason explain, why are alloys commonly used to make elements of electrical heating devices. { } $\displaystyle 1190$-$\displaystyle 1$
(i)
The given electric circuit is a part of an electrical device. Use the information given in the electric circuit diagram to calculate :
(I)
Potential difference across the ends of resistor $\displaystyle \mathrm{R}_{2}$.
(II)
Value of resistor $\displaystyle \mathrm{R}_{2}$.
(III)
Value of resistor $\displaystyle \mathrm{R}_{1}$.
(ii)
Write the factors on which resistance of a conductor depends and derive the formula for resistance of a given conductor.
(B)
How much electric current will an electric iron draw from $\displaystyle 220$ V source if the resistance of its heating element when hot, is $\displaystyle 55 \Omega$ ? Calculate the power consumed by the electric iron when it is operated at $\displaystyle 220$ V.
(ii)
In a house, $\displaystyle 3$ bulbs of $\displaystyle 100$ watt 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)
With reason explain, why are alloys commonly used to make elements of electrical heating devices. { } $\displaystyle 1190$-$\displaystyle 1$
Marking-scheme solution
Let I be the total current flowing through the circuit and
$\displaystyle \mathrm{I}_{1}$ and $\displaystyle \mathrm{I}_{2}$ be the currents flowing through $\displaystyle 4 \Omega(\mathrm{R})$ and $\displaystyle \mathrm{R}_{2}$ resistors.
(i)
(I)
Potential difference across $\displaystyle \mathrm{R}_{2}$ is same as that of across $\displaystyle 4 \Omega$ resistor as they are connected in parallel.
\[\begin{aligned}
\therefore \quad \mathrm{V}\left(\operatorname{across} \mathrm{R}_{2}\right) & =\mathrm{I}_{1} \mathrm{R} \\
& =1.5 \times 4 \\
& =6 \mathrm{~V}
\end{aligned}
\]
(II)
Current flowing through $\displaystyle \mathrm{I}_{2}=\mathrm{I}-\mathrm{I}_{1}$
\[=2.0-1.5=0.5 \mathrm{~A}
\]
\[\begin{array}{l}
\mathrm{R}_{2}=\frac{\mathrm{V}}{\mathrm{I}_{2}} \\
R_{2}=\frac{6}{0.5} \\
\mathrm{R}_{2}=12 \Omega
\end{array}
\]
(III)
Potential difference across $\displaystyle 2 \Omega$ resistor
\[\begin{array}{l}
\mathrm{V}=\mathrm{IR} \\
\mathrm{~V}=2 \times 2=4 \mathrm{~V}
\end{array}
\]
Potential difference across $\displaystyle \mathrm{R}_{1}=12-(6+4)=2 \mathrm{~V}$
\[\begin{array}{l}
R_{1}=\frac{V}{I} \\
R_{1}=\frac{2}{2} \Rightarrow \mathrm{R}_{1}=1 \Omega
\end{array}
\]
(ii)
Resistance of conductor depends on
Length of the conductor / R $\displaystyle \alpha l$
Area of cross section / $\displaystyle \mathrm{R} \alpha \frac{1}{\mathrm{~A}}$
R $\displaystyle \alpha l$
\[\begin{array}{l}
\mathrm{R} \alpha \frac{1}{\mathrm{~A}} \\
\mathrm{R} \alpha \frac{l}{\mathrm{~A}} \\
\mathrm{R}=\rho \frac{l}{\mathrm{~A}}
\end{array}
\]
Where $\displaystyle \rho=$ Resistivity (a proportionality constant)
PAGE $\displaystyle 11$ \{$\displaystyle 31$-$\displaystyle 4$-$\displaystyle 1$\}
(B)
\[\begin{array}{l}
I=\frac{V}{R} \\
I=\frac{220}{55} \\
I=4 \mathrm{~A}
\end{array}
\]
Power of electric iron $\displaystyle \mathrm{P}=\mathrm{VI}$
\[\begin{array}{l}
\mathrm{P}=220 \times 4 \\
\mathrm{P}=880 \mathrm{~W}
\end{array}
\]
(ii)
\[\mathrm{E}=\mathrm{P} \times \mathrm{t}
\]
\[\begin{aligned}
\text { Energy (3 bulbs) } & =3 \times 100 \times 5 \\
& =1500 \\
& =1.5 \mathrm{kWh} \\
\text { Energy (electric heater) } & =1.0 \times 0.5 \\
& =0.5 \mathrm{kWh}
\end{aligned}
\]
Total energy consumed ($\displaystyle 1$ day) $\displaystyle =1.5+0.5=2 \mathrm{kWh}$
Total energy consumed ($\displaystyle 30$ days) $\displaystyle =30 \times 2$
\[\begin{aligned}
& =60 \mathrm{kWh} \\
& =60 \mathrm{units} \\
\text { Total cost }= & \text { Units × Rate } \\
= & 60 \times 3.60 \\
= & ₹ 216
\end{aligned}
\]
(iii)
The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures.
PAGE $\displaystyle 12$ \{$\displaystyle 31$-$\displaystyle 4$-$\displaystyle 1$\}
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