3. The equivalent resistance of 5 equal resistors connected in parallel is \( 20 \Omega \), then their equivalent resistance when they connected in series... \( \begin{array}{llll}\text { A }-300 \Omega & \text { B }-500 \Omega & \text { C- } 270 \Omega & \text { D-540 } \Omega\end{array} \)

To solve this problem, we need to use the formulas for equivalent resistance in parallel and series. First, let's find the resistance of each individual resistor when they are connected in parallel. Given: - The equivalent resistance of 5 equal resistors connected in parallel is \( 20 \Omega \). The formula for the equivalent resistance of resistors in parallel is: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \] Since all resistors are equal, we can denote the resistance of each resistor as \( R \). Therefore, the formula becomes: \[ \frac{1}{20} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \] \[ \frac{1}{20} = \frac{5}{R} \] Now, solve for \( R \): \[ R = \frac{5}{\frac{1}{20}} \] \[ R = 5 \times 20 \] \[ R = 100 \Omega \] So, each resistor has a resistance of \( 100 \Omega \). Now, let's find the equivalent resistance when these 5 resistors are connected in series. The formula for the equivalent resistance of resistors in series is: \[ R_{eq} = R_1 + R_2 + R_3 + R_4 + R_5 \] Since all resistors are equal, the formula becomes: \[ R_{eq} = 100 + 100 + 100 + 100 + 100 \] \[ R_{eq} = 500 \Omega \] Therefore, the equivalent resistance when the 5 resistors are connected in series is \( 500 \Omega \). The correct answer is: \( \text{B } - 500 \Omega \)