Three resistors with values of 5Ω, 10Ω and 15Ω are linked in series, and the resulting connections are connected to the battery with a 30 V voltage. The circuit is set up to include an ammeter and a voltmeter. Draw a connection to all the devices in the proper correct order. What are the current flowing and potential differences across 10 Ω resistance?

The potential difference across a 10 Ω the resistor is 10 𝑣𝑜𝑙𝑡𝑠. 

Given that, three resistors are connected in series. 

𝑅₁ = 5 Ω 

𝑅₂ = 10 Ω 

𝑅₃ = 15 Ω 

Voltage 𝑉 = 30 𝑉 

We need to calculate the resistance 

R = 5 + 10 + 15 = 30 Ω 

We need to calculate the current in the circuit 

Using ohm's law, 

The current is directly proportional to the voltage. 

𝑉 = 𝐼𝑅 

Where 𝐼 = current 

𝑉 = voltage 

𝑅 = resistance 

Put the value into the formula 

30 = I $\times$30

I = 1 A

The current through a 10 Ω the resistor will remain the same, equal to the total current in the circuit because the resistor is connected in series.

The potential difference across 10 $\Omega$ resistance,

V = IR

V = 1 $\times$10

V = 10 V

Hence, the potential difference across a 10$\Omega$ resistor is 10 volts.