Question

A 5.00 Ω, a10.0 Ω, and a 15.0 Ω resistor are connected in series across a 90.0v battery. What is the voltage drop across the 5.00 Ω resistor?

Answer 1

`V_(1) = 15\,V`

Step-by-step 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:

`V_(batt) = V_(1) + V_(2) + V_(3)`

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:

`V_(1) = \left((5\,\Omega)/(30\,\Omega) \right)\cdot (90\,V)`

`V_(1) = 15\,V`

Answer 2

15volts

Step-by-step 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