Question

$B_1,B_2$ and $B_3$ are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
What happens to the reading of $A_1,A_2,A_3$ and A when the bulb $B_2$ gets fused?

Answer 1

Hint : Try to find the resistance of individual bulb and then use Ohm's Law to get current in each arm.

Step 1 : Find resistance of individual bulb.

Given: Three bulbs connected in parallel to a source of $4.5V$ and net current in the circuit is $3A$

Assume every bulb has a resistance $R$

Net resistance of all three together in parallel (case $1$)

$\frac{1}{R_{net1}}=$$\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$R_1=R_2=R_3=R$

$⟹$ $R_{net1}=\frac{R}{3}$

Current in whole circuit = $3A$(Given).

Using Ohm's Law

$V=I{R}_{net}$

$4.5=3\times \frac{R}{3}$

$\therefore$ $R=4.5\Omega$

Step 2 : Find the net resistance and current in the circuit when bulb $B_2$ is fused.

Now, as $B_2$ is fused$B_1$ and $B_3$ are in parallel (case $2$)

$\therefore$ $\frac{1}{R_{net2}}=$ $\frac{1}{R}+\frac{1}{R}$

$R_{net2}=$ $\frac{R}{2}=$$2.25\Omega$

$\therefore$ Current in the circuit $I=\frac{V}{R_{net2}}=\frac{4.5}{2.25}=$ $2A$

Reading of ammeter $A=2A$

Step 3 : Find current in individual branches

As $B_2$ is fused no current flows through $B_2$ $\therefore$ $A_2=0A$

The net current of $2A$ divides equally in all two branches as the resistance of both the bulbs is same.

$A_1=1A$ and $A_2=1A$

Final Step : When Bulb $B_2$ fuses,

The reading of $A_1$ becomes $1A,$

The reading of $A_2$ becomes $0A,$

The reading of $A_3$ becomes $1A,$

The reading of $A$ becomes $2A,$