krit.club logo

Electricity - Series and Parallel Circuits

Grade 9IB

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

🔑Concepts

In a Series Circuit, there is only one path for the electrons to flow. This means the current II remains constant at every point in the circuit: Itotal=I1=I2=I3I_{total} = I_1 = I_2 = I_3.

In a Parallel Circuit, the current divides into different branches. The sum of the currents in each branch equals the total current entering the junction: Itotal=I1+I2++InI_{total} = I_1 + I_2 + \dots + I_n.

The Potential Difference (Voltage) in a series circuit is shared between components. The sum of the voltages across individual resistors equals the source voltage: Vtotal=V1+V2++VnV_{total} = V_1 + V_2 + \dots + V_n.

In a parallel circuit, the Potential Difference across each branch is identical to the source voltage: Vtotal=V1=V2=V3V_{total} = V_1 = V_2 = V_3.

Resistance in Series: Adding more resistors in series increases the total resistance RsR_s, as the electrons must flow through more obstacles in a single path.

Resistance in Parallel: Adding more resistors in parallel decreases the total resistance RpR_p, as it provides more paths for the current to flow. The total resistance is always less than the smallest individual resistor: 1Rp=1Rn\frac{1}{R_p} = \sum \frac{1}{R_n}.

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points: V=IRV = IR.

📐Formulae

V=I×RV = I \times R

Rseries=R1+R2+R3++RnR_{series} = R_1 + R_2 + R_3 + \dots + R_n

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

P=V×I=I2R=V2RP = V \times I = I^2R = \frac{V^2}{R}

💡Examples

Problem 1:

Calculate the total resistance and the total current for a circuit with a 12V12V battery connected to two resistors, 4Ω4\Omega and 8Ω8\Omega, in series.

Solution:

Rtotal=4Ω+8Ω=12ΩR_{total} = 4\Omega + 8\Omega = 12\Omega. Total current I=VRtotal=12V12Ω=1.0AI = \frac{V}{R_{total}} = \frac{12V}{12\Omega} = 1.0A.

Explanation:

In a series circuit, resistors are simply added together. Using Ohm's Law I=VRI = \frac{V}{R}, we find the current flowing through the entire loop.

Problem 2:

Two resistors, R1=10ΩR_1 = 10\Omega and R2=10ΩR_2 = 10\Omega, are connected in parallel to a 5V5V power supply. Determine the equivalent resistance RpR_p and the current through R1R_1.

Solution:

1Rp=110Ω+110Ω=210ΩRp=5Ω\frac{1}{R_p} = \frac{1}{10\Omega} + \frac{1}{10\Omega} = \frac{2}{10\Omega} \Rightarrow R_p = 5\Omega. Current through R1R_1: I1=VR1=5V10Ω=0.5AI_1 = \frac{V}{R_1} = \frac{5V}{10\Omega} = 0.5A.

Explanation:

For parallel circuits, the reciprocal sum is used for resistance. The voltage across each resistor is the same as the source (5V5V), allowing us to calculate individual branch currents using Ohm's Law.

Series and Parallel Circuits - Revision Notes & Key Formulas | IB Grade 9 Science