Question

Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.

Answer 1

Answer:

I = 0.25 A

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 :

$R=R_1+R_2+R_3+....$

So,

$R=15+15+15+25+25\\\\=95\ \Omega$

The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.

V = IR

$I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A$

So, the current of 0.25 A passes through each bulb.