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Physics: Electricity and Magnetism - Current, Voltage, and Resistance

Grade 8IB

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

🔑Concepts

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Electric Current (II): The rate of flow of electric charge (QQ) through a conductor. It is measured in Amperes (AA) using an ammeter connected in series.

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Potential Difference or Voltage (VV): The amount of energy or work done (WW) per unit charge to move it between two points in a circuit. It is measured in Volts (VV) using a voltmeter connected in parallel.

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Resistance (RR): The opposition offered by a substance to the flow of electric current. It is measured in Ohms (Ω\Omega). Factors affecting resistance include length, cross-sectional area, material, and temperature.

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Ohm's Law: At a constant temperature, the current (II) flowing through a conductor is directly proportional to the potential difference (VV) across its ends, expressed as V=I×RV = I \times R.

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Series Circuit: Components are connected end-to-end. The current remains the same through all components (Itotal=I1=I2I_{total} = I_1 = I_2), but the total voltage is shared (Vtotal=V1+V2V_{total} = V_1 + V_2).

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Parallel Circuit: Components are connected across the same two points. The voltage remains the same across all branches (Vtotal=V1=V2V_{total} = V_1 = V_2), but the total current is the sum of currents in each branch (Itotal=I1+I2I_{total} = I_1 + I_2).

📐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

💡Examples

Problem 1:

A charge of 30 C30\ C passes through a light bulb in 1010 seconds. Calculate the current flowing through the bulb.

Solution:

Given: Q=30 CQ = 30\ C, t=10 st = 10\ s. Using the formula I=QtI = \frac{Q}{t}, we get I=3010=3 AI = \frac{30}{10} = 3\ A.

Explanation:

Current is defined as the rate of flow of charge. Dividing the total charge by the time taken gives the current in Amperes.

Problem 2:

A resistor is connected to a 12 V12\ V battery, and a current of 0.5 A0.5\ A flows through it. What is the resistance of the resistor?

Solution:

Given: V=12 VV = 12\ V, I=0.5 AI = 0.5\ A. According to Ohm's Law, R=VIR = \frac{V}{I}. Substituting the values: R=120.5=24 ΩR = \frac{12}{0.5} = 24\ \Omega.

Explanation:

Resistance is the ratio of potential difference to current. Using Ohm's Law, we rearrange the formula to solve for RR.

Problem 3:

Calculate the total resistance of two resistors, 4 Ω4\ \Omega and 6 Ω6\ \Omega, when they are connected in series.

Solution:

For a series circuit, Rtotal=R1+R2R_{total} = R_1 + R_2. Substituting the values: 4+610\begin{array}{r} 4 \\ + 6 \\ \hline 10 \end{array} So, Rtotal=10 ΩR_{total} = 10\ \Omega.

Explanation:

In a series connection, the total resistance is simply the sum of the individual resistances because the current has only one path to flow through.