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Physics - Electricity (Current, Voltage, and Resistance)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electric current (II) is defined as the rate of flow of electric charge (QQ). It is measured in Amperes (AA).

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Potential Difference or Voltage (VV) is the work done (WW) or energy transferred per unit charge passing between two points. It is measured in Volts (VV).

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Resistance (RR) is the opposition to the flow of current. It is measured in Ohms (Ω\Omega).

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Ohm’s Law states that the current flowing through a metallic conductor is directly proportional to the potential difference across it, provided that temperature and other physical conditions remain constant (V=IRV = IR).

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In a series circuit, the current (II) is the same at all points, while the total voltage is the sum of voltages across each component: Vtotal=V1+V2+…V_{total} = V_1 + V_2 + \dots.

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In a parallel circuit, the potential difference (VV) across each branch is the same, while the total current is the sum of the currents in each branch: Itotal=I1+I2+…I_{total} = I_1 + I_2 + \dots.

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The power (PP) dissipated in an electrical component is the product of the potential difference and the current, measured in Watts (WW).

📐Formulae

I=QtI = \frac{Q}{t}

V=WQV = \frac{W}{Q}

V=I×RV = I \times R

Rseries=R1+R2+R3+…R_{series} = R_1 + R_2 + R_3 + \dots

1Rparallel=1R1+1R2+1R3+…\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

P=V×IP = V \times I

E=P×t=V×I×tE = P \times t = V \times I \times t

💡Examples

Problem 1:

A charge of 30 C30 \text{ C} passes through a light bulb in 1 minute1 \text{ minute}. Calculate the current flowing through the bulb.

Solution:

I=Qt=30 C60 s=0.5 AI = \frac{Q}{t} = \frac{30 \text{ C}}{60 \text{ s}} = 0.5 \text{ A}

Explanation:

To find current, we divide the total charge by the time in seconds. Note that 1 minute1 \text{ minute} must be converted to 60 seconds60 \text{ seconds} to keep units consistent with S.I. standards.

Problem 2:

A resistor has a potential difference of 12 V12 \text{ V} across it and a current of 3 A3 \text{ A} flowing through it. Calculate its resistance.

Solution:

R=VI=12 V3 A=4 ΩR = \frac{V}{I} = \frac{12 \text{ V}}{3 \text{ A}} = 4 \text{ } \Omega

Explanation:

Using Ohm's Law, we rearrange V=IRV = IR to solve for RR by dividing voltage by current.

Problem 3:

Two resistors, R1=6 ΩR_1 = 6 \text{ } \Omega and R2=3 ΩR_2 = 3 \text{ } \Omega, are connected in parallel. Calculate the total equivalent resistance of the circuit.

Solution:

1Rtotal=16+13=16+26=36  ⟹  Rtotal=63=2 Ω\frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} \implies R_{total} = \frac{6}{3} = 2 \text{ } \Omega

Explanation:

For parallel circuits, we sum the reciprocals of the individual resistances. After finding the sum, we must take the reciprocal again to find the final resistance value.