Question

If two bulbs, whose resistance are in the ratio of $1:2$, are connected in series. The power dissipated in them has the ratio of :

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:4$

Answer 1

The correct option is B $1:2$

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.
The power dissipated is given as $P=VI.$ That is,$P=I\times I\times R$ (From ohm's law)
When the resistance is increased in the second bulb, the power dissipated becomes $P^{\prime }=I\times I\times 2R.$

That is, $P^{\prime }=2P.$ Hence, the power dissipated in the two bulbs has the ratio of $1:2.$