Since V and R of both the bulbs B2 and B3 remain the same even if bulb B, gets fused so glow of B2 and B3 remain the same. Since bulbs are identical, so their resistance is equal (i.e. resistance of each bulb = RΩ). 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 B2 gets fused, only two bulbs B1 and B2 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 B1 and B3 are in parallel and have same resistance, so 2A current will be equally distributed between B1 and B3. Therefore, reading of ammeter A1 = 1A Reading of ammeter A3 = 1A Circuit containing B2 is broken, so no current flows through this circuit. Hence reading of ammeter A2 = zero.
- Power dissipated in the circuit,
P = V × I
= 4.5 × 3 = 13.5W
