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Physics - Current Electricity

Grade 10ICSE

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

🔑Concepts

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Electric Current (II): It is defined as the rate of flow of electric charge through any cross-section of a conductor. I=QtI = \frac{Q}{t}, measured in Amperes (AA).

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Potential Difference (VV): The work done in moving a unit positive charge from one point to another in an electric circuit. V=WQV = \frac{W}{Q}, measured in Volts (VV).

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Ohm's Law: At a constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. V=IRV = IR.

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Resistance (RR): The obstruction offered by the conductor to the flow of current. It depends on length (ll), area of cross-section (AA), material, and temperature. R=ρlAR = \rho \frac{l}{A}.

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Specific Resistance (Resistivity) (ρ\rho): The resistance of a conductor of unit length and unit area of cross-section. It is a characteristic property of the material and does not change with dimensions.

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Electromotive Force (EMF) (EE): The potential difference across the terminals of a cell when no current is drawn from it (open circuit).

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Internal Resistance (rr): The resistance offered by the electrolyte inside the cell to the flow of current. Terminal voltage V=E−IrV = E - Ir.

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Electrical Energy (WW): The total work done by the source of electricity in maintaining the current in a circuit. W=VIt=I2RtW = VIt = I^2Rt.

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Electrical Power (PP): The rate at which electrical energy is consumed. P=VI=I2R=V2RP = VI = I^2R = \frac{V^2}{R}. Unit: Watt (WW).

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Kilowatt-hour (kWhkWh): The commercial unit of electrical energy. 1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}.

📐Formulae

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

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

V=IRV = IR

R=ρlAR = \rho \frac{l}{A}

Rs=R1+R2+R3+…R_s = R_1 + R_2 + R_3 + \dots

1Rp=1R1+1R2+1R3+…\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

E=V+v=I(R+r)E = V + v = I(R + r)

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

W=P×t=VItW = P \times t = VIt

Total Cost=Energy in kWh×Rate per unit\text{Total Cost} = \text{Energy in kWh} \times \text{Rate per unit}

💡Examples

Problem 1:

A wire of resistance 9Ω9 \Omega is tripled on itself (folded into three equal parts). Calculate its new resistance.

Solution:

Let the original length be ll and area be AA. R=ρlA=9ΩR = \rho \frac{l}{A} = 9 \Omega. When tripled on itself, the new length l′=l3l' = \frac{l}{3} and the new area A′=3AA' = 3A. New resistance R′=ρl′A′=ρl/33A=19(ρlA)=19×9=1ΩR' = \rho \frac{l'}{A'} = \rho \frac{l/3}{3A} = \frac{1}{9} (\rho \frac{l}{A}) = \frac{1}{9} \times 9 = 1 \Omega.

Explanation:

Folding a wire decreases its length and increases its cross-sectional area, both of which contribute to a reduction in total resistance.

Problem 2:

A cell of EMF 2.0V2.0 V and internal resistance 1.2Ω1.2 \Omega is connected to an external resistor of 3.8Ω3.8 \Omega. Calculate the current in the circuit and the terminal voltage.

Solution:

Total resistance Rtotal=R+r=3.8Ω+1.2Ω=5.0ΩR_{total} = R + r = 3.8 \Omega + 1.2 \Omega = 5.0 \Omega. Current I=ER+r=2.05.0=0.4AI = \frac{E}{R+r} = \frac{2.0}{5.0} = 0.4 A. Terminal voltage V=IR=0.4×3.8=1.52VV = IR = 0.4 \times 3.8 = 1.52 V.

Explanation:

The terminal voltage is lower than the EMF because of the 'lost volts' (v=Irv = Ir) across the internal resistance of the cell.

Problem 3:

Calculate the electrical energy consumed by a 100W100 W bulb used for 55 hours daily in a month of 3030 days, in kWhkWh.

Solution:

Power P=100W=0.1kWP = 100 W = 0.1 kW. Time t=5 hours/day×30 days=150 hourst = 5 \text{ hours/day} \times 30 \text{ days} = 150 \text{ hours}. Energy E=P×t=0.1kW×150h=15kWhE = P \times t = 0.1 kW \times 150 h = 15 kWh.

Explanation:

To find energy in commercial units (kWhkWh), convert power to kilowatts and time to hours before multiplying.