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Electricity — Science STD 10 — Question

CBSE BoardEnglish MediumSTD 10ScienceElectricity5 Marks
Question
$B_1, B_2$ and $B_3$ are three identical bulbs connected as shown in Figure. When all the three bulbs glow, a current of 3 A is recorded by the ammeter A
.
i. What happens to the glow of the other two bulbs when the bulb $B_1$ gets fused?
ii. What happens to the reading of $A_1, A_2, A_3$ and $A$ when the bulb $B_2$ gets fused?
iii. How much power is dissipated in the circuit when all the three bulbs glow together?
✓

Answer

i. The glow of bulb depends upon the energy disspated per second i.e. $P =\frac{ V ^2}{ R }$. Since V and R of both the bulbs $B_2$ and $B_3$ remain the same even if bulb $B$, gets fused so glow of $B_2$ and $B_3$ remain the same.
ii. Since bulbs are identical, so their resistance is equal (i.e. resistance of each bulb $=R \Omega$ ).
When all bulbs glow, net resistnace of the circuit is given by,
$\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}$
Or, $R ^{\prime}=\frac{ R }{3}$
$I=3 A, V=4.5 V$
Using, $V= R ^{\prime}$, we get
$4.5=3 \times \frac{R}{3} \text { or } R=4.5 \Omega$
When $B _2$ gets fused, only two bulbs $B _1$ and $B _2$ in parallel are in the circuit.
$\therefore$ Net resistance of the circuit is given by,
$\frac{1}{R}=\frac{1}{4.5}+\frac{1}{4.5}=\frac{2}{4.5} \text { or } R=\frac{4.5}{2} \Omega$
$\therefore I=\frac{V}{R}=\frac{4.5 \times 2}{4.5}=2 A$
Thus, reading of ammeter $A=2 A$
Since $B_1$ and $B_3$ are in parallel and have same resistance, so $2 A$ current will be equally distributed between $B_1$ and $B_3$. Therefore, reading of ammeter $A_1=1 A$ Reading of ammeter $A_3=1 A$ Circuit containing $B_2$ is broken, so no current flows through this circuit. Hence reading of ammeter $A_2=$ zero.
iii. Power dissipated in the circuit,
$P=V \times I$
$=4.5 \times 3=13.5 W$

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