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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.

Respuesta :

hi
Resistor 5+9+13
Volts 24
V=RXI
[tex]25 \div 27 \\[/tex]

Answer:

Current, I = 0.89 A

Explanation:

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

[tex]R_s=R_1+R_2+R_3[/tex]

[tex]R_s=5+9+13=27\ \Omega[/tex]

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 :

[tex]I=\dfrac{V}{R_s}[/tex]

[tex]I=\dfrac{24}{27}[/tex]  

I = 0.89 A

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

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