In Thanaweya Amma physics, most students know the formulas. What costs marks is choosing one that does not fit the situation. A formula is a statement about a particular setup, so read the setup first: what is connected to what, what is given, what is asked, and in which units. The three examples below are original practice problems on electricity topics from the grade 12 book.
The routine
- 1List the given values with their units, and the unknown.
- 2Draw the circuit or situation if the question does not.
- 3Name the physical situation (closed circuit with internal resistance, parallel branches, shunt...) and only then the formula.
- 4Convert units (mA to A, cm to m) before substituting.
- 5Check: does the answer have the right unit and a sensible size?
Example 1: a battery with internal resistance
A battery of emf VB = 12 V and internal resistance r = 1 Ω is connected to an external resistance R = 5 Ω. Find the current and the potential difference across R. Situation: closed circuit, so Ohm's law for a closed circuit: I = VB / (R + r) = 12 / 6 = 2 A. Then V = IR = 2 × 5 = 10 V. Check: V = VB − Ir = 12 − 2 = 10 V.
Common mistake: using I = V/R with 12 V and 5 Ω, which gives 2.4 A, because it ignores the internal resistance that the question gave you.
Example 2: resistors in parallel
Resistors of 6 Ω and 3 Ω are connected in parallel to a 12 V source (internal resistance neglected). Find the total current and the current in each. Situation: parallel, so the voltage is the same across both. 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω and I = 12 / 2 = 6 A. Branches: 12/6 = 2 A and 12/3 = 4 A. Check: 2 + 4 = 6 A, as Kirchhoff's first law requires.
Common mistake: adding the resistances (9 Ω) as if they were in series. In parallel, the total resistance is always smaller than the smallest branch.
Example 3: turning a galvanometer into an ammeter
A galvanometer of resistance Rg = 50 Ω gives full-scale deflection at Ig = 2 mA. What shunt makes it read up to 1 A? Situation: the shunt is in parallel, so it has the same potential difference as the galvanometer: Ig·Rg = (I − Ig)·Rs. Convert: Ig = 0.002 A. Rs = (0.002 × 50) / (1 − 0.002) = 0.1 / 0.998 ≈ 0.1 Ω. Check: a small shunt is expected, because most of the current must bypass the coil.
Common mistake: substituting 2 instead of 0.002 because the current was given in milliamperes.
Check that a formula applies to the situation before you put numbers into it.
How EXAMEV helps
The EXAMEV 3rd Secondary Physics Test Bank (Egypt) has practice questions lesson by lesson, from electric current, Ohm's and Kirchhoff's laws to magnetic fields, measuring instruments and induction, with an explanation for every answer. The examples here are EXAMEV practice, not official exam questions; check the ministry's current specification for what is examined this year. For a full-year timetable, read the Thanaweya Amma study plan on this blog.