CBSE 2025 · Region 4 · Set 2 · Q21 · 2 marks
In the given figure, three identical bulbs $\displaystyle \mathrm{P}, \mathrm{Q}$ and S are connected to a battery.
(i)Compare the brightness of bulbs P and Q with that of bulb S when key K is closed.(ii)Compare the brightness of the bulbs S and Q when the key K is opened. Justify your answer in both cases.Two cells of emf $\displaystyle 10$ V each, two resistors of $\displaystyle 20 \Omega$ and $\displaystyle 10 \Omega$ and a bulb B of $\displaystyle 10 \Omega$ resistance are connected together as shown in the figure. Find the current that flows through the bulb.
In the given figure, three identical bulbs $\displaystyle \mathrm{P}, \mathrm{Q}$ and S are connected to a battery.
(i)
Compare the brightness of bulbs P and Q with that of bulb S when key K is closed.
(ii)
Compare the brightness of the bulbs S and Q when the key K is opened. Justify your answer in both cases.
Two cells of emf $\displaystyle 10$ V each, two resistors of $\displaystyle 20 \Omega$ and $\displaystyle 10 \Omega$ and a bulb B of $\displaystyle 10 \Omega$ resistance are connected together as shown in the figure. Find the current that flows through the bulb.
Marking-scheme solution
(i)
Brightness of the bulb S will be more than bulbs P and Q.
The current flowing through the bulb S is twice the current in bulbs P and Q.
Brightness of the bulbs S and Q will be the same.
The current flowing through both bulbs is the same.
Brightness of the bulb S will be more than bulbs P and Q.
The potential difference across S is twice the potential difference across bulbs P and Q.
Brightness of both bulbs S and Q is the same.
The potential difference across S and Q will be the same.
By applying Kirchhoff's loop rule to closed loops ABCFA and FCDEF:
$\displaystyle 2 I_{1}-3 I_{2}=1$ ----($\displaystyle 1$)
$\displaystyle I_{1}+I_{2}=1$ -----($\displaystyle 2$)
On solving, current through the bulb:
$\displaystyle I_{2}=\frac{1}{5} \mathrm{~A}$
Current ElectricityKirchhoff’s RulesApplyvery_short_answermedium
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