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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.

Electricity (Current, Voltage, and Resistance) Revision - Grade 9 Science IGCSE