Middle School

We appreciate your visit to A 5 00 Ω a 10 0 Ω and a 15 0 Ω resistor are connected in series across a 90 0 V battery What. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in series across a 90.0 V battery.

What is the voltage drop across the 5.00 Ω resistor?

Answer :

Answer:

[tex]V_{1} = 15\,V[/tex]

Explanation:

The configuration in series means that voltage of the battery is equal to the sum of the voltage drop of the resistors, which are modelled Ohm's Law:

[tex]V_{batt} = V_{1} + V_{2} + V_{3}[/tex]

Where voltage drop is directly proportional to electrical resistance. Hence, the voltage drop across the 5 Ω is determined by a simple rule of three:

[tex]V_{1} = \left(\frac{5\,\Omega}{30\,\Omega} \right)\cdot (90\,V)[/tex]

[tex]V_{1} = 15\,V[/tex]

Thanks for taking the time to read A 5 00 Ω a 10 0 Ω and a 15 0 Ω resistor are connected in series across a 90 0 V battery What. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada

Answer:

15volts

Explanation:

According to ohms law, the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends. Mathematically,

E = IRt

E is the source voltage

I is the total current

Rt is the effective resistance.

Before we can get the voltage drip across a load, we must know the current flowing through it.

Given E = 90V

Rt = 5+10+15 (for a series connection)

Rt = 30Ω

I = E/Rt

I = 90/30

I = 3A

Note that In a series connected circuit, same current but different voltage flows through the circuit.

Therefore the current in the 5Ω resistor is also 3A.

To get the voltage drip across the 5Ω resistor we use the formula

V=IR

V = 3(5)

V = 15volts