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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 .
  1. What happens to the glow of the other two bulbs when the bulb gets fused?
  2. What happens to the reading of $A_1, A_2, A_3$ and $A$ when the bulb $B_2$ gets fused?
  3. 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,

$\frac{1}{\text{R}'}=\frac{1}{\text{R}}+\frac{1}{\text{R}}+\frac{1}{\text{R}}=\frac{3}{\text{R}}$

Or, $\text{R}'=\frac{\text{R}}{3}$

I = 3A, V = 4.5V

Using, V = IR', we get

$4.5=3\times\frac{\text{R}}{3}\ \text{or R }=4.5\Omega$

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,

$\frac{1}{\text{R}}=\frac{1}{4.5}+\frac{1}{4.5}=\frac{2}{4.5}\ \text{or R }=\frac{4.5}{2}\Omega$

$\therefore\ \text{I}=\frac{\text{V}}{\text{R}}=\frac{4.5\times2}{4.5}=2\text{A}$

Thus, reading of ammeter A = 2A

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.
  1. Power dissipated in the circuit,
P = V × I

= 4.5 × 3 = 13.5W

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