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Imagine an electric kettle drawing 10 amps from a 120-volt wall outlet to boil water for three minutes. How much actual electrical charge passed through the heating coil, and how many joules of energy were transferred?

To solve problems like this on your exam, you only need six core formulas and the skill to link them together.

Every quantitative circuit problem links six fundamental variables: **charge** ($Q$), **current** ($I$), **potential difference** or voltage ($V$), **resistance** ($R$), **power** ($P$), and **energy** ($E$).

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In 1827, German physicist Georg Ohm discovered that current through a conductor is directly proportional to voltage, giving us **Ohm's law** ($V = IR$).

Always verify that your quantities are in base **SI units** before calculating: time in seconds ($s$), current in amperes ($A$), and charge in coulombs ($C$).

What happens when an exam question doesn't give you the variable you need for a direct formula?

When you don't have a direct equation, you can substitute one formula into another to eliminate unknown terms.

For example, power is $P = IV$; substituting Ohm's law ($V = IR$) gives $P = I(IR) = I^2 R$, while substituting $I = \frac{V}{R}$ gives $P = \frac{V^2}{R}$.

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Similarly, substituting $Q = It$ into the energy formula $E = QV$ yields $E = (It)V = VIt$, which is identical to multiplying power ($P = IV$) by time ($t$).

Let's see how this two-step substitution strategy works on a complete exam-style calculation.

## Worked Example: Multi-Step Circuit Problem

**Problem:** A $24\,\Omega$ heating element is connected to a $12\text{ V}$ battery for $5.0\text{ minutes}$. Calculate the total charge transferred and the total thermal energy released.

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<html lang="en">
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<body>

  
  <div class="header-bar">
    <div class="given-chips">
      <span class="chip v-v">V = 12 V</span>
      <span class="chip v-r">R = 24 Ω</span>
      <span class="chip v-t">t = 5 min</span>
    </div>
    <div class="nav-controls">
      <button class="nav-btn active" onclick="setStep(0)">1</button>
      <button class="nav-btn" onclick="setStep(1)">2</button>
      <button class="nav-btn" onclick="setStep(2)">3</button>
      <button class="play-btn" id="playBtn" onclick="togglePlay()" title="Play/Pause">⏸</button>
    </div>
  </div>

  
  <div class="steps-container">
    
    
    <div class="step-card active" id="card-0" onclick="setStep(0)">
      <div class="step-header">
        <span class="step-tag" style="color:#0284c7;">Step 1: Current</span>
        <span class="step-formula">I = V ÷ R</span>
      </div>
      <div class="calc-row">
        <div class="math-expr">
          <span class="chip v-t">5 min × 60 = 300 s</span>
          <span class="carry-arrow">→</span>
          <span class="chip v-i">I</span>
          <span>=</span>
          <span class="fraction">
            <span class="fraction-top chip v-v">12 V</span>
            <span class="fraction-bot chip v-r">24 Ω</span>
          </span>
        </div>
        <div class="ans-box v-i">
          <span class="check">✓</span> 0.50 A
        </div>
      </div>
    </div>

    
    <div class="step-card" id="card-1" onclick="setStep(1)">
      <div class="step-header">
        <span class="step-tag" style="color:#059669;">Step 2: Charge</span>
        <span class="step-formula">Q = I × t</span>
      </div>
      <div class="calc-row">
        <div class="math-expr">
          <span class="chip v-q">Q</span>
          <span>=</span>
          <span class="chip v-i">0.50 A</span>
          <span>×</span>
          <span class="chip v-t">300 s</span>
        </div>
        <div class="ans-box v-q">
          <span class="check">✓</span> 150 C
        </div>
      </div>
    </div>

    
    <div class="step-card" id="card-2" onclick="setStep(2)">
      <div class="step-header">
        <span class="step-tag" style="color:#db2777;">Step 3: Energy</span>
        <span class="step-formula">E = Q × V</span>
      </div>
      <div class="calc-row">
        <div class="math-expr">
          <span class="chip v-e">E</span>
          <span>=</span>
          <span class="chip v-q">150 C</span>
          <span>×</span>
          <span class="chip v-v">12 V</span>
        </div>
        <div class="ans-box v-e">
          <span class="check">✓</span> 1800 J
        </div>
      </div>
    </div>

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First, convert time into seconds: $5.0\text{ min} \times 60\text{ s/min} = 300\text{ s}$. Then find current using $I = \frac{V}{R} = \frac{12\text{ V}}{24\,\Omega} = 0.50\text{ A}$.

Next, compute charge using $Q = It = 0.50\text{ A} \times 300\text{ s} = 150\text{ C}$, and total energy using $E = QV = 150\text{ C} \times 12\text{ V} = 1800\text{ J}$ (or $1.8\text{ kJ}$).

Why do so many students lose easy marks on questions like this even when their algebra is correct?

The most frequent error is calculating power using $P = I^2 R$ by forgetting to square the current, or squaring both $I$ and $R$ instead of just $I$.

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <meta name="viewport" content="width=device-width, initial-scale=1.0">
  <meta name="aspect-ratio" content="4:3">
  <style>
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      box-sizing: border-box;
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<body>

  <div class="header-bar">
    <span>Given Values:</span>
    <span class="vals">I = 0.50 A, &nbsp;R = 24 &Omega;</span>
  </div>

  <div class="grid">
    
    <div class="panel panel-error">
      <div class="title title-error">❌ Common Squaring Errors</div>

      
      <div class="calc-card">
        <div class="math strike">P = (I &times; R)&sup2; = (12)&sup2; = 144 W</div>
        <span class="badge badge-error">Squared both I and R</span>
      </div>

      
      <div class="calc-card">
        <div class="math strike">P = I &times; R = 0.50 &times; 24 = 12 W</div>
        <span class="badge badge-error">Forgot power of 2</span>
      </div>
    </div>

    
    <div class="panel panel-correct">
      <div class="title title-correct">✅ Correct Method (<span class="math">P = I&sup2;R</span>)</div>

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        <div class="step-row">
          <span class="step-num">1</span>
          <span class="math">P = <span class="highlight-i">(0.50)&sup2;</span> &times; 24</span>
        </div>
        <span class="badge badge-tip" style="margin-left: 22px;">Only I is squared</span>

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          <span class="step-num">2</span>
          <span class="math">P = 0.25 &times; 24</span>
        </div>

        <div class="result-box math">
          P = 6.0 W
        </div>
      </div>
    </div>
  </div>

</body>
</html>

Another major trap is leaving time in minutes or hours; electrical equations require seconds because $1\text{ Ampere} = 1\text{ Coulomb per second}$.

When solving circuit questions, list all given values with their standard units first, identify the target variable, and map your substitution path before touching your calculator.

science

Mastering GCSE Electricity Equation Substitutions. Select, rearrange, and combine the key electrical formulas: Q=It, V=IR, P=IV, P=I²R, E=Pt, and E=QV.

Electric Circuits and Quantitative Equation Substitutions (KS4 / GCSE)

GCSE Masterclass: Electrical Power, Energy Transfers, and Circuit Calculations

GCSE Physics: Electrical Power, Work Done, and Equation Rearrangements

Pedagogical best practice for KS4 multi-equation mastery relies on explicit 'faded scaffolding': start with single-equation substitutions, progress to linked two-step substitutions ($P=IV$ with $V=IR$), and conclude with unprompted multi-step word problems requiring unit conversion and formula derivation. Emphasise dimensional analysis and checking solutions using alternate formula paths (e.g. verifying $E=Pt$ against $E=QV$).

Mastering GCSE Electricity Equation Substitutions

The Six Core Electrical Equations

Combining Equations by Substitution