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Electricity and Magnetism - Electric circuits (Series and parallel)

Grade 11IGCSEPhysics

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

🔑Concepts

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In a Series Circuit, there is only one path for the electric current to flow. If the circuit is broken at any point, the current stops flowing everywhere.

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The Current (II) in a series circuit is the same at every point: Itotal=I1=I2=I3I_{total} = I_1 = I_2 = I_3.

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The Total Potential Difference (VtotalV_{total}) in a series circuit is shared between the components: Vtotal=V1+V2+...+VnV_{total} = V_1 + V_2 + ... + V_n.

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In a Parallel Circuit, there are multiple paths (branches) for the current. If one branch is broken, current can still flow through the others.

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The Potential Difference (VV) across each branch in a parallel circuit is the same: Vtotal=V1=V2=V3V_{total} = V_1 = V_2 = V_3.

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The Total Current (ItotalI_{total}) in a parallel circuit is the sum of the currents in each branch: Itotal=I1+I2+...+InI_{total} = I_1 + I_2 + ... + I_n.

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The Total Resistance (RR) in a series circuit increases as more resistors are added: Rtotal=R1+R2+...R_{total} = R_1 + R_2 + ....

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The Total Resistance (RR) in a parallel circuit decreases as more branches are added. The total resistance is always less than the resistance of the smallest individual resistor.

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An Ammeter measures current and must always be connected in series. It has very low resistance.

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A Voltmeter measures potential difference and must always be connected in parallel across the component. It has very high resistance.

📐Formulae

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} + ...

Rparallel=R1×R2R1+R2 (for two resistors only)R_{parallel} = \frac{R_1 \times R_2}{R_1 + R_2} \text{ (for two resistors only)}

P=I×VP = I \times V

E=I×V×tE = I \times V \times t

💡Examples

Problem 1:

Two resistors, R1=5 ΩR_1 = 5\, \Omega and R2=10 ΩR_2 = 10\, \Omega, are connected in series to a 15 V15\, \text{V} battery. Calculate the total current flowing through the circuit.

Solution:

Rtotal=R1+R2=5 Ω+10 Ω=15 ΩR_{total} = R_1 + R_2 = 5\, \Omega + 10\, \Omega = 15\, \Omega I=VRtotal=15 V15 Ω=1.0 AI = \frac{V}{R_{total}} = \frac{15\, \text{V}}{15\, \Omega} = 1.0\, \text{A}

Explanation:

First, find the total resistance by adding the individual resistances in series. Then, use Ohm's Law (V=IRV = IR) to solve for the current.

Problem 2:

Two resistors, R1=6 ΩR_1 = 6\, \Omega and R2=3 ΩR_2 = 3\, \Omega, are connected in parallel. What is the total resistance of this combination?

Solution:

1Rtotal=1R1+1R2\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} 1Rtotal=16+13=16+26=36\frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} Rtotal=63=2.0 ΩR_{total} = \frac{6}{3} = 2.0\, \Omega

Explanation:

For parallel circuits, use the reciprocal formula. After summing the fractions, remember to take the reciprocal of the result to find RtotalR_{total}. Note that 2.0 Ω2.0\, \Omega is less than both 3 Ω3\, \Omega and 6 Ω6\, \Omega.

Problem 3:

In a parallel circuit with a 12 V12\, \text{V} supply, branch A has a 4 Ω4\, \Omega resistor and branch B has a 6 Ω6\, \Omega resistor. Calculate the current in branch A.

Solution:

Vbranch=Vsupply=12 VV_{branch} = V_{supply} = 12\, \text{V} IA=VRA=12 V4 Ω=3.0 AI_A = \frac{V}{R_A} = \frac{12\, \text{V}}{4\, \Omega} = 3.0\, \text{A}

Explanation:

In a parallel circuit, the voltage across each branch is equal to the source voltage. We can apply Ohm's Law directly to the specific branch.

Electric circuits (Series and parallel) Revision - Grade 11 Physics IGCSE