If you have a 90 v power source with a 30-ω and 60-ω resistor in series, what is the voltage drop over the 60-ω resistor?

30 v
2 v
60 v
90 v

= 60 V

Explanation:

Total resistance = 30 Ω+ 60 Ω

                           = 90 Ω

Voltage = 90 V

We can calculate current through the circuit.

I = V/R

 = 90 v/90Ω

 = 1 A

Since the resistors are in series arrangement, then the current through both resistors are the same.

Current through 60Ω resistor is 1 A

Therefore; V = IR

                     = 1 A× 60 Ω

                     = 60 V

Therefore;  the voltage drop over the 60-Ω resistor is 60 V

an energy transformation flow diagram is shown

where is it ?