Electricity — Questions & Answers

Sample questions from Electricity

1. Q1. A 5 Ω and a 10 Ω resistor are connected in series across a 15 V battery. The potential difference across the 10 Ω resistor is:

   • A.10 V✓
   • B.5 V
   • C.15 V
   • D.1.5 V
   Solution

   Total R = 15 Ω, so I = 15/15 = 1 A and V = 1 × 10 = 10 V. The full 15 V would appear only across the whole combination.

2. Q2. The relation V = IR is known as:

   • A.Ohm’s law✓
   • B.Joule’s law
   • C.Newton’s law
   • D.Faraday’s law
   Solution

   Ohm’s law relates potential difference, current and resistance.

3. Q3. In the given circuit diagram, resistors of 2 Ω, 3 Ω and 5 Ω are connected in series with a 10 V battery and an ammeter. The ammeter reading is:

   • A.0.5 A
   • B.1 A✓
   • C.2 A
   • D.10 A
   Solution

   Rs = 2 + 3 + 5 = 10 Ω, so I = V/Rs = 10/10 = 1 A. The same current flows through every element of a series circuit.

4. Q4. A 6 Ω and a 3 Ω resistor are connected in parallel. The equivalent resistance is:

   • A.2 Ω✓
   • B.9 Ω
   • C.4.5 Ω
   • D.18 Ω
   Solution

   A is correct: 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω, which is less than either of them, as a parallel combination always is. B adds them as though they were in series. C takes the average of the two. D multiplies them without dividing by their sum.

5. Q5. A 4 Ω and a 12 Ω resistor are joined in parallel across a 6 V battery. The equivalent resistance and the total current are:

   • A.16 Ω and 0.375 A
   • B.8 Ω and 0.75 A
   • C.3 Ω and 2 A✓
   • D.3 Ω and 0.5 A
   Solution

   C is correct: 1/R = 1/4 + 1/12 = 1/3, so R = 3 Ω, and the total current is I = 6/3 = 2 A, which agrees with the branch currents 6/4 = 1.5 A and 6/12 = 0.5 A. A uses the series rule. B averages the two resistances. D has the right resistance but divides the voltage wrongly.

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