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Electricity - Define electric current, potential difference, and state Ohm's law

Grade 10CBSE

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

🔑Concepts

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Electric Current (II) is the rate of flow of electric charges (electrons) through a conductor. It is measured in Amperes (AA) using an Ammeter, which is always connected in series.

A simple circuit diagram showing a battery, an ammeter in series, and a resistor.
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Electric Potential Difference (VV) between two points is the work done to move a unit positive charge from one point to the other. It is measured in Volts (VV) using a Voltmeter connected in parallel.

Circuit showing a voltmeter connected in parallel across a resistor.
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Ohm's Law states that the current (II) flowing through a conductor is directly proportional to the potential difference (VV) across its ends, provided temperature remains constant: V∝IV \propto I or V=IRV = IR.

A resistor labeled with points A and B showing current direction to illustrate Ohm's Law.
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Resistance (RR) is the property of a conductor to resist the flow of charges. Its SI unit is Ohm (Ω\Omega). 1 Ohm1 \text{ Ohm} is the resistance of a conductor such that a potential difference of 1 V1 \text{ V} causes a current of 1 A1 \text{ A} to flow.

📐Formulae

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

Q=n⋅eQ = n \cdot e

V=WQV = \frac{W}{Q}

1 A=1 C1 s1 \text{ A} = \frac{1 \text{ C}}{1 \text{ s}}

1 V=1 J1 C1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}

💡Examples

Problem 1:

A current of 0.5 A0.5 \text{ A} is drawn by a filament of an electric bulb for 1010 minutes. Find the amount of electric charge that flows through the circuit.

Solution:

Given: I=0.5 AI = 0.5 \text{ A}, t=10 minutes=10×60=600 st = 10 \text{ minutes} = 10 \times 60 = 600 \text{ s}. Using Q=I×tQ = I \times t: Q=0.5 A×600 s=300 CQ = 0.5 \text{ A} \times 600 \text{ s} = 300 \text{ C}.

Explanation:

To find the charge, we multiply the current by the time in seconds. 1010 minutes must be converted to 600600 seconds for standard SI units.

Problem 2:

How much work is done in moving a charge of 2 C2 \text{ C} across two points having a potential difference of 12 V12 \text{ V}?

Solution:

Given: Q=2 CQ = 2 \text{ C}, V=12 VV = 12 \text{ V}. Using W=V×QW = V \times Q: W=12 V×2 C=24 JW = 12 \text{ V} \times 2 \text{ C} = 24 \text{ J}.

Explanation:

Work done is the product of the potential difference and the amount of charge moved.

Problem 3:

Calculate the number of electrons constituting one Coulomb of charge.

Solution:

Given: Q=1 CQ = 1 \text{ C}, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}. Using n=Qen = \frac{Q}{e}: n=11.6×10−19≈6.25×1018n = \frac{1}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{18} electrons.

Explanation:

By applying the quantization of charge formula Q=neQ = ne, we find the total number of elementary charges required to sum up to one Coulomb.

Problem 4:

The potential difference between the terminals of an electric heater is 60 V60 \text{ V} when it draws a current of 4 A4 \text{ A} from the source. What current will the heater draw if the potential difference is increased to 120 V120 \text{ V}?

Circuit diagram for the heater example with a 120V source and 15 Ohm resistance.

Solution:

Step 1: Find Resistance (RR) using the first case. Given V1=60 VV_1 = 60 \text{ V}, I1=4 AI_1 = 4 \text{ A}. According to Ohm's law, R=V1I1R = \frac{V_1}{I_1} R=60 V4 A=15 ΩR = \frac{60 \text{ V}}{4 \text{ A}} = 15 \text{ } \Omega

Step 2: Find new current (I2I_2) when V2=120 VV_2 = 120 \text{ V}. I2=V2RI_2 = \frac{V_2}{R} I2=120 V15 Ω=8 AI_2 = \frac{120 \text{ V}}{15 \text{ } \Omega} = 8 \text{ A} The current through the heater becomes 8 A8 \text{ A}.

Explanation:

Resistance remains constant for the same heater. Since VV is doubled (60 V60 \text{ V} to 120 V120 \text{ V}), the current II also doubles (4 A4 \text{ A} to 8 A8 \text{ A}) following V∝IV \propto I.

Problem 5:

Resistance of an electric iron is 50 Ω50 \text{ } \Omega. If a current of 4.4 A4.4 \text{ A} flows through it, calculate the potential difference between the two terminals of the electric iron.

Circuit segment showing a 50 Ohm resistor with 4.4 Amps of current flowing through it.

Solution:

Given: Resistance R=50 ΩR = 50 \text{ } \Omega Current I=4.4 AI = 4.4 \text{ A}

Using Ohm's Law: V=I×RV = I \times R V=4.4 A×50 ΩV = 4.4 \text{ A} \times 50 \text{ } \Omega V=220 VV = 220 \text{ V}

The potential difference is 220 V220 \text{ V}.

Explanation:

By multiplying the flow of charge (current) by the opposition to that flow (resistance), we find the electrical pressure (potential difference) required.