⚡ Chapter 16: Electricity – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 16 Electricity including current, resistance, series and parallel circuits, e.m.f., power, energy, and cost calculations. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 16: Electricity
Electricity covers electric current, Ohm's law, resistance, series and parallel circuits, electric power, energy, and household wiring. Includes solved examples and numerical problems.
📐 Numerical Problems & Solutions (PECTAA 2026)
16.1 Current from Charge and Time
A charge of 25 coulombs passes through a circuit in 10 seconds. Calculate the current flowing through the circuit.
Given Data
\(Charge\)\(= Q = 25\ C\)
\(Time\)\(= t = 10\ s\)
To Find
\(Current = I = ?\)
Solution
By using formula of electric current:
\[\begin{aligned}
I &= \frac{Q}{t} \\
I &= \frac{25}{10} \\
\boldsymbol{I} &= \boldsymbol{2.5\ A}
\end{aligned}\]
✅ \(I = 2.5\ A\)
16.2 Series Resistance and Current
Three resistors of \(4\Omega\), \(6\Omega\), and \(8\Omega\) are connected in series to a \(24V\) battery. Find the total resistance and the current flowing through the circuit.
Given Data
\(R_1\)\(= 4\ \Omega\)
\(R_2\)\(= 6\ \Omega\)
\(R_3\)\(= 8\ \Omega\)
\(Voltage\)\(= V = 24\ V\)
To Find
\(Total\ resistance = R_e = ?\)
\(Current = I = ?\)
Solution
By using the formula for series combination:
\[\begin{aligned}
R_e &= R_1 + R_2 + R_3 \\
R_e &= 4 + 6 + 8 \\
\boldsymbol{R_e} &= \boldsymbol{18\ \Omega}
\end{aligned}\]
By using Ohm's law:
\[\begin{aligned}
V &= IR_e \\
I &= \frac{V}{R_e} \\
I &= \frac{24}{18} \\
\boldsymbol{I} &= \boldsymbol{1.33\ A}
\end{aligned}\]
✅ \(R_e = 18\ \Omega\) | \(I = 1.33\ A\)
16.3 Parallel Resistance and Current
Two resistors of \(10\Omega\) and \(30\Omega\) are connected in parallel across a \(20V\) power supply. Calculate the total resistance and the current drawn from the power supply.
Given Data
\(R_1\)\(= 10\ \Omega\)
\(R_2\)\(= 30\ \Omega\)
\(Voltage\)\(= V = 20\ V\)
To Find
\(Total\ resistance = R_e = ?\)
\(Current = I = ?\)
Solution
By using the formula for parallel combination:
\[\begin{aligned}
\frac{1}{R_e} &= \frac{1}{R_1} + \frac{1}{R_2} \\
\frac{1}{R_e} &= \frac{1}{10} + \frac{1}{30} \\
\frac{1}{R_e} &= \frac{3 + 1}{30} \\
\frac{1}{R_e} &= \frac{4}{30} \\
R_e &= \frac{30}{4} \\
\boldsymbol{R_e} &= \boldsymbol{7.5\ \Omega}
\end{aligned}\]
By using Ohm's law:
\[\begin{aligned}
V &= IR_e \\
I &= \frac{V}{R_e} \\
I &= \frac{20}{7.5} \\
\boldsymbol{I} &= \boldsymbol{2.67\ A}
\end{aligned}\]
✅ \(R_e = 7.5\ \Omega\) | \(I = 2.67\ A\)
16.4 Total e.m.f. in Series
Four 1.5 V batteries are connected in series in a flashlight. Determine the total electromotive force (e.m.f.) supplied to the circuit.
Given Data
\(E_1 = E_2 = E_3 = E_4\)\(= 1.5\ V\)
To Find
\(Total\ e.m.f. = E_{total} = ?\)
Solution
When multiple voltage sources are connected in series, their total electromotive force is the sum of their individual e.m.f.
\[\begin{aligned}
E_{total} &= E_1 + E_2 + E_3 + E_4 \\
E_{total} &= 1.5 + 1.5 + 1.5 + 1.5 \\
\boldsymbol{E_{total}} &= \boldsymbol{6\ V}
\end{aligned}\]
✅ \(E_{total} = 6\ V\)
16.5 Power and Energy from Current and Resistance
A circuit has a 5 A current passing through a \(7\Omega\) resistor. Calculate the power dissipated in the resistor and the energy consumed in 10 minutes.
Given Data
\(Current\)\(= I = 5\ A\)
\(Resistance\)\(= R = 7\ \Omega\)
\(Time\)\(= t = 10\ min = 600\ s\)
To Find
\(Power = P = ?\)
\(Energy = E = ?\)
Solution
By using formula of power:
\[\begin{aligned}
P &= I^2 R \\
P &= (5)^2 (7) \\
P &= (25)(7) \\
\boldsymbol{P} &= \boldsymbol{175\ W}
\end{aligned}\]
Using the formula for energy:
\[\begin{aligned}
E &= Pt \\
E &= (175)(600) \\
E &= 105000\ J \\
\boldsymbol{E} &= \boldsymbol{105\ kJ}
\end{aligned}\]
✅ \(P = 175\ W\) | \(E = 105\ kJ\)
📘 Examples with Solutions
Ex 16.1 Potential Difference from Work and Charge
A battery does 120 J of work to move a charge of \(20C\) through a circuit. What is the potential difference across the circuit?
Given Data
\(Work\)\(= W = 120\ J\)
\(Charge\)\(= Q = 20\ C\)
To Find
\(Potential\ difference = V = ?\)
Solution
By using the formula for potential difference:
\[\begin{aligned}
V &= \frac{W}{Q} \\
V &= \frac{120}{20} \\
\boldsymbol{V} &= \boldsymbol{6\ V}
\end{aligned}\]
✅ \(V = 6\ V\)
Ex 16.2 Parallel Resistance and Current
Two resistors of \(8\Omega\) and \(24\Omega\) are connected in parallel across a \(30V\) power supply. Calculate the total resistance and the current drawn from the power supply.
Given Data
\(R_1\)\(= 8\ \Omega\)
\(R_2\)\(= 24\ \Omega\)
\(Voltage\)\(= V = 30\ V\)
To Find
\(Total\ resistance = R_e = ?\)
\(Current = I = ?\)
Solution
By using the formula for parallel combination:
\[\begin{aligned}
\frac{1}{R_e} &= \frac{1}{R_1} + \frac{1}{R_2} \\
\frac{1}{R_e} &= \frac{1}{8} + \frac{1}{24} \\
\frac{1}{R_e} &= \frac{3 + 1}{24} \\
\frac{1}{R_e} &= \frac{4}{24} \\
R_e &= \frac{24}{4} \\
\boldsymbol{R_e} &= \boldsymbol{6\ \Omega}
\end{aligned}\]
By using Ohm's law:
\[\begin{aligned}
V &= IR_e \\
I &= \frac{V}{R_e} \\
I &= \frac{30}{6} \\
\boldsymbol{I} &= \boldsymbol{5\ A}
\end{aligned}\]
✅ \(R_e = 6\ \Omega\) | \(I = 5\ A\)
Ex 16.3 Resistance from Resistivity
An aluminium wire has a length of \(15m\), a cross-sectional area of \(3\times 10^{-6}m^2\), and a resistivity of \(2.82\times 10^{-8}\Omega m\). Calculate its resistance.
Given Data
\(Length\)\(= L = 15\ m\)
\(Area\)\(= A = 3\times 10^{-6}\ m^2\)
\(Resistivity\)\(= \rho = 2.82\times 10^{-8}\ \Omega m\)
To Find
\(Resistance = R = ?\)
Solution
By using the formula for resistance:
\[\begin{aligned}
R &= \rho \frac{L}{A} \\
R &= \frac{(2.82\times 10^{-8})(15)}{3\times 10^{-6}} \\
\boldsymbol{R} &= \boldsymbol{0.141\ \Omega}
\end{aligned}\]
✅ \(R = 0.141\ \Omega\)
Ex 16.4 Monthly Electricity Cost
Calculate the one-month cost of using a 100 W ceiling fan for 10 hours daily in your room. Assume that the price of one unit is Rs. 25.
Given Data
\(Power\)\(= P = 100\ W\)
\(Time\ per\ day\)\(= t = 10\ h\)
\(Unit\ price\)\(= Rs.\ 25\)
To Find
\(Monthly\ cost\ (30\ days) = ?\)
Solution
Using the formula for electricity cost:
\[\begin{aligned}
Monthly\ cost &= \frac{P(W) \times t(h) \times Price \times 30}{1000} \\
&= \frac{100 \times 10 \times 25 \times 30}{1000} \\
\boldsymbol{Cost} &= \boldsymbol{Rs.\ 750}
\end{aligned}\]
✅ Cost = Rs. 750
📐 Key Formulas – Electricity
Electric Current: \( I = \frac{Q}{t} \)
Ohm's Law: \( V = IR \)
Series Resistance: \( R_e = R_1 + R_2 + R_3 \)
Parallel Resistance: \( \frac{1}{R_e} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
Electric Power: \( P = IV = I^2R = \frac{V^2}{R} \)
Electrical Energy: \( E = Pt \)
Resistivity: \( R = \rho \frac{L}{A} \)
Electricity Cost: \( Cost = \frac{P(W) \times t(h) \times Price}{1000} \)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13Sound
Unit 14Light
Unit 15Electrostatics
Unit 16✨ Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and conversions. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus