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Electricity - Current, Voltage, and Resistance

Grade 9IB

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) through a conductor, measured in Amperes (AA). One Ampere is equal to one Coulomb per second (1A=1C/s1 A = 1 C/s).

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Potential Difference or Voltage (VV) is the work done or energy transferred (EE) per unit charge (QQ) as it moves between two points in a circuit, measured in Volts (VV).

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Resistance (RR) is the property of a material that opposes the flow of electric current, measured in Ohms (Ω\Omega). It depends on the material, length, cross-sectional area, and temperature.

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Ohm's Law states that for an ohmic conductor at a constant temperature, the current (II) passing through it is directly proportional to the potential difference (VV) across it, represented as V=IRV = IR.

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In a Series Circuit, the current (II) remains constant through all components, while the total voltage is the sum of individual voltages (Vtotal=V1+V2+...V_{total} = V_1 + V_2 + ...).

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In a Parallel Circuit, the potential difference (VV) is the same across all branches, while the total current is the sum of the currents in each branch (Itotal=I1+I2+...I_{total} = I_1 + I_2 + ...).

📐Formulae

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

V=EQV = \frac{E}{Q}

V=I×RV = I \times R

Rseries=R1+R2+R3+...R_{series} = R_1 + R_2 + R_3 + ...

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

P=V×IP = V \times I

💡Examples

Problem 1:

A resistor has a resistance of 15Ω15 \Omega. If a potential difference of 45V45 V is applied across it, calculate the current flowing through the resistor.

Solution:

I=VR=45V15Ω=3AI = \frac{V}{R} = \frac{45 V}{15 \Omega} = 3 A

Explanation:

Using Ohm's Law, we rearrange V=IRV = IR to solve for II. Substituting the given values for voltage and resistance gives the current in Amperes.

Problem 2:

Calculate the total resistance of a circuit where two resistors, R1=4ΩR_1 = 4 \Omega and R2=6ΩR_2 = 6 \Omega, are connected in parallel.

Solution:

1Rtotal=14+16=3+212=512Ω−1⇒Rtotal=125=2.4Ω\frac{1}{R_{total}} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \Omega^{-1} \Rightarrow R_{total} = \frac{12}{5} = 2.4 \Omega

Explanation:

For parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances. After adding the fractions, we take the reciprocal to find RtotalR_{total}.

Problem 3:

How much charge passes through a light bulb in 22 minutes if the current is 0.5A0.5 A?

Solution:

Q=I×t=0.5A×(2×60s)=60CQ = I \times t = 0.5 A \times (2 \times 60 s) = 60 C

Explanation:

First, convert time from minutes to seconds (2×60=120s2 \times 60 = 120 s). Then, apply the formula Q=ItQ = It to find the total charge in Coulombs (CC).