CBSE 2025 · Region 5 · Set 3 · Q22 · 2 marks
Determine the maximum and minimum resistance which can be obtained by joining five resistors of $\displaystyle \frac{1}{5} \Omega$ each.Calculate potential difference across a $\displaystyle 4 \Omega$ resistor that produces $\displaystyle 100$ J of heat every second.
Determine the maximum and minimum resistance which can be obtained by joining five resistors of $\displaystyle \frac{1}{5} \Omega$ each.
Calculate potential difference across a $\displaystyle 4 \Omega$ resistor that produces $\displaystyle 100$ J of heat every second.
Marking-scheme solution
The maximum resistance is obtained when resistors are connected in series combination.
Thus \(\displaystyle \mathrm{R}_{\mathrm{S}}=5 \times \frac{1}{5} \Omega=1 \Omega\)
The minimum resistance is obtained when resistors are connected in parallel combination.
\[\begin{aligned}
& \frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\dfrac{1}{5} \Omega}+\frac{1}{\dfrac{1}{5} \Omega}+\frac{1}{\dfrac{1}{5} \Omega}+\frac{1}{\dfrac{1}{5} \Omega}+\frac{1}{\dfrac{1}{5} \Omega} \\
& \frac{1}{\mathrm{R}_{\mathrm{p}}}=(5+5+5+5+5) \\
& \mathrm{R}_{\mathrm{p}}=\frac{1}{25} \Omega
\end{aligned}
\]
Heat, \(\displaystyle \mathrm{H}=100 \mathrm{~J}\)
\[\begin{aligned}
& \mathrm{R}=4 \Omega \\
& \mathrm{t}=1 \mathrm{~s} \\
& \mathrm{H}=\frac{V^{2} t}{R} \\
& \Rightarrow \mathrm{~V}=\sqrt{\frac{H R}{t}}=\sqrt{\frac{400}{1}}=20 \mathrm{~V}
\end{aligned}
\]
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