Question

Bulb A rated 160 W, 40 V and Bulb B rated 40 W, 40 V are connected as shown in the diagram.
(a) Calculate the ratio V\(_1\) : V\(_2\)
(b) If the bulb A fuses, the current in the circuit remains the same. State True or False.

Hint:

In a series circuit, voltage divides in the ratio of the resistances (\(V \propto R\)), while current is constant. In a parallel circuit, current divides in the inverse ratio of the resistances (\(I \propto 1/R\)), while voltage is constant.

Answer 1

Step 1: Calculate the resistance of each bulb.
The resistance (\(R\)) of a bulb can be calculated from its power (\(P\)) and voltage (\(V\)) rating using the formula \(P = V^2 / R\), so \(R = V^2 / P\). - For Bulb A: \(R_A = \frac{(40 \text{ V})^2}{160 \text{ W}} = \frac{1600}{160} = 10 \, \Omega\) - For Bulb B: \(R_B = \frac{(40 \text{ V})^2}{40 \text{ W}} = \frac{1600}{40} = 40 \, \Omega\)
(a) Calculate the ratio V\(_1\) : V\(_2\):
The bulbs A and B are connected in series. In a series circuit, the same current (\(I\)) flows through all components. The voltage across each component is given by Ohm's law, \(V = IR\). - Voltage across A: \(V_1 = I \times R_A\) - Voltage across B: \(V_2 = I \times R_B\) The ratio of the voltages is: \[ \frac{V_1}{V_2} = \frac{I \times R_A}{I \times R_B} = \frac{R_A}{R_B} \] \[ \frac{V_1}{V_2} = \frac{10 \, \Omega}{40 \, \Omega} = \frac{1}{4} \] So, the ratio \(V_1 : V_2\) is 1 : 4.
(b) True or False:
The statement is False. The bulbs are connected in a series circuit. If bulb A fuses, its filament breaks, creating an open circuit. In an open circuit, no current can flow. Therefore, the current in the circuit will drop to zero, not remain the same.