CBSE 2025 · Region 5 · Set 2 · Q22 · 2 marks
Use Ohm's law to determine the potential difference across the $\displaystyle 6 \Omega$ resistor in the following circuit when key is closed :
Prove that if the current through a resistor is increased by $\displaystyle 100$%, then the increase in power dissipated through the resistor will be $\displaystyle 300$%.
Use Ohm's law to determine the potential difference across the $\displaystyle 6 \Omega$ resistor in the following circuit when key is closed :
Prove that if the current through a resistor is increased by $\displaystyle 100$%, then the increase in power dissipated through the resistor will be $\displaystyle 300$%.
Marking-scheme solution
(A) Total resistance in the circuit
\[\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}=12 \Omega
\]
Voltage of the battery = $\displaystyle 6$ V
\[\therefore I=\frac{V}{R}=\frac{6}{12}=0.5 \mathrm{~A}
\]∴ Potential difference across \(\displaystyle 6 \Omega\) resistor \(\displaystyle =0.5 \mathrm{~A} \times 6 \Omega=3.0 \mathrm{~V}\) OR
(B) \(\displaystyle \mathrm{P}_{1}=\mathrm{I}^{2} \mathrm{R} \mathrm{P}_{2}=(2 \mathrm{I})^{2} \mathrm{R}=4 \mathrm{I}^{2} \mathrm{R}[100 \%\) increase in current means current becomes 2I]
∴ Increase in power dissipated \(\displaystyle =\mathrm{P}_{2}-\mathrm{P}_{1}=4 \mathrm{I}^{2} \mathrm{R}-\mathrm{I}^{2} \mathrm{R}=3 \mathrm{I}^{2} \mathrm{R}\)
\[=3 \mathrm{P}_{1}
\]
Percentage increase in power dissipated \(\displaystyle =\frac{3 \mathrm{P}_{1}}{\mathrm{P}_{1}} \times 100=300 \%\)ElectricityOhm's LawApplyvery_short_answermedium
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CBSE Class 10 Science past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.