Answer:
I = 0.25 A
Step-by-step explanation:
Given that,
Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.
In series combination, the equivalent resistance is given by :
![R=R_1+R_2+R_3+....](https://img.qammunity.org/2022/formulas/physics/college/r6y95a9tjtmn056j8wu4nei43z2g1vehaj.png)
So,
![R=15+15+15+25+25\\\\=95\ \Omega](https://img.qammunity.org/2022/formulas/physics/college/gttqqy007ks346v8b42enowy4tpq4vde89.png)
The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.
V = IR
![I=(V)/(R)\\\\I=(24)/(95)\\\\I=0.25\ A](https://img.qammunity.org/2022/formulas/physics/college/1npz3f3mb4eo4d8u51zqgapj5s76q6mlpj.png)
So, the current of 0.25 A passes through each bulb.