Two resistors, each of 10 Ω are connected in parallel. The combination is connected in series to a third resistor of 10 Ω. The equivalent resistance of this arrangement is: 

The correct answer is 15 Ω.

Key Points

• Two resistors of 10 Ω each connected in parallel have an equivalent resistance of $\frac{10 \times 10}{10 + 10} = 5 \Omega$ .
• The equivalent resistance of the parallel combination is 5 Ω.
• This 5 Ω resistor is connected in series with a third resistor of 10 Ω.
• In a series connection, the total resistance is the sum of individual resistances.
• Therefore, the equivalent resistance of the entire arrangement is $5 \Omega + 10 \Omega = 15 \Omega$ .

Additional Information

• Series and Parallel Circuits
  • In a series circuit, the current flows through each component in sequence, and the total resistance is the sum of individual resistances.
  • In a parallel circuit, the voltage across each component is the same, and the total resistance is less than the smallest individual resistance.
• Ohm's Law
  • Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
  • It is mathematically represented as $V = IR$ , where V is voltage, I is current, and R is resistance.
• Combination of Resistors
  • Resistors can be combined in series or parallel to achieve desired resistance values in a circuit.
  • The method of combination affects the overall resistance and the behavior of the circuit.