On a model railroad, a 15 Ω track-side resistor needs a current of 0.30 A through it. How much voltage must be across the resistor to drive that current?
Ohm's law is the same equation as always: I = V/R. This time the unknown is the voltage V, so make V the subject.
Multiply both sides by R. The R on the right cancels, leaving I × R = V. Turn it around: V = IR.
The three solving steps stay exactly the same: write down the values and mark the unknown, select the correct equation, substitute in the values and simplify.
The railroad's 15 Ω resistor must carry 0.30 A. Calculate the voltage across it.
Write down the values in the question:
I = 0.30 A
R = 15 Ω
V = ?
Write down the equation:
I = V/R
Make V the subject:
V = IR
Substitute in the values:
V = (0.30)(15)
Simplify:
V = 4.5 V
A 25 Ω resistor carries 0.40 A. Enter the voltage across it in V.
A 60 Ω lamp carries 0.20 A. Enter the voltage across it in V.
The railroad's point motor has a coil of resistance 36 Ω and carries 250 mA. Calculate the voltage across the coil.
Write down the values in the question:
I = 250 mA
R = 36 Ω
V = ?
Convert the current to amperes:
I = 250/1000 A = 0.25 A
Write down the equation:
V = IR
Substitute in the values:
V = (0.25)(36)
Simplify:
V = 9.0 V
Four more. Work each one on paper first; the rearranged equation V = IR goes straight into step 2.
A 40 Ω coil carries 150 mA. Enter the voltage across it in V.
A 22 Ω resistor carries 0.50 A. Enter the voltage across it in V.
A 16 Ω heating element carries 2.0 A. Enter the voltage across it in V.
A 30 Ω resistor carries 0.25 A. Enter the voltage across it in V.
A 20 Ω speaker coil carries 0.35 A. Enter the voltage across the coil in V.
A relay coil of resistance 60 Ω carries 450 mA. Enter the voltage across it in V.
In the circuit shown, what does the voltmeter read? Enter it in V.