Question

Given that a 5.0 Ω resistor, a 9.0 Ω resistor, and a 13.0 Ω resistor are connected in series with a 24.0 V battery, calculate the current in each resistor.

Answer 1

hi
Resistor 5+9+13
Volts 24
V=RXI

`25 / 27 \\`

Answer 2

Current, I = 0.89 A

Step-by-step explanation:

It is given that, three resistors are connected in series. Their equivalent is given by :

`R_s=R_1+R_2+R_3`

`R_s=5+9+13=27\ \Omega`

Voltage of battery, V = 24 V

In series combination the current flowing across each resistor is same. The current can be calculated using Ohm's law :

`I=(V)/(R_s)`

`I=(24)/(27)`

I = 0.89 A

So, the current flowing in each resistor is 0.89 A. Hence, this is the required solution.