10.12-10.14 - Resistance and circuit rules

10.12-10.14 - Resistance and circuit rules

Use resistance to control current, calculate with V = IR and explain why series and parallel arrangements change total resistance differently.

Resistance and current control

Resistance describes how strongly a component opposes the flow of charge. At a fixed potential difference, a larger resistance gives a smaller current, while a smaller resistance gives a larger current. This is a change in how much charge passes each second, not a claim that current is a substance moving more slowly.

A variable resistor lets a person change resistance in a circuit. Moving its slider or turning its control changes the length of resistive material included in the conducting route. When it is connected in series with another component:

  • increasing the variable resistance increases the circuit's resistance and decreases the current;
  • decreasing the variable resistance decreases the circuit's resistance and increases the current.

The control does not adjust itself like a sensor. The user deliberately sets it to change or limit the current.

Potential difference VV, current II and resistance RR are connected by:

V=I×RV = I \times R
  • VV is potential difference in volts, V
  • II is current in amperes, A
  • RR is resistance in ohms, Ω\Omega

This relationship is included in the supplied equation support; practise choosing it, using consistent units and rearranging it.

For a component, the potential difference must be measured across it and the current must be the current through it. The resistance can then be found by rearranging:

R=VIR = \frac{V}{I}

Worked example

A resistor has a potential difference of 6.0 V6.0\text{ V} across it and a current of 250 mA250\text{ mA} through it. Calculate its resistance.

Convert the current to amperes before substitution:

250 mA=2501000 A=0.250 A250\text{ mA} = \frac{250}{1000}\text{ A} = 0.250\text{ A}

Then rearrange and substitute:

R=VI=6.0 V0.250 A=24 ΩR = \frac{V}{I} = \frac{6.0\text{ V}}{0.250\text{ A}} = 24\ \Omega

The result is reported to two significant figures, matching the potential difference. A sense-check using V=IRV = IR gives 0.250×24=6.0 V0.250 \times 24 = 6.0\text{ V}, so the value is consistent.

At the same 6.0 V6.0\text{ V}, doubling the resistance from 12 Ω12\ \Omega to 24 Ω24\ \Omega changes the current from 0.50 A0.50\text{ A} to 0.25 A0.25\text{ A}. The potential difference is fixed, so greater resistance means less charge passes per second.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account