A battery of 6 V is connected with a series combination of five resistors of \( 0.1 \Omega, 0.2 \Omega, 0.3 \Omega, 0.4 \Omega \) and \( 0.5 \Omega \). How much current would flow through the \( 0.3 \Omega \) resistor ? Justify your answer.

To find the current flowing through the \( 0.3 \Omega \) resistor in a series circuit, we can follow these steps: 1. **Calculate the total resistance in the circuit**: In a series circuit, the total resistance \( R_t \) is the sum of all individual resistances. \[ R_t = R_1 + R_2 + R_3 + R_4 + R_5 \] where \( R_1 = 0.1 \Omega \), \( R_2 = 0.2 \Omega \), \( R_3 = 0.3 \Omega \), \( R_4 = 0.4 \Omega \), and \( R_5 = 0.5 \Omega \). 2. **Use Ohm's Law to find the current**: Ohm's Law states that \( V = I \cdot R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. Rearranging this gives us: \[ I = \frac{V}{R_t} \] where \( V = 6 \, V \). Now, let's perform the calculations step by step. ### Step 1: Calculate Total Resistance \[ R_t = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 \] Calculating this: \[ R_t = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5 \, \Omega \] ### Step 2: Calculate Current Using Ohm's Law Now, substituting the values into Ohm's Law: \[ I = \frac{6}{1.5} \] Calculating this: \[ I = 4 \, A \] ### Conclusion Since the resistors are in series, the same current flows through all resistors, including the \( 0.3 \Omega \) resistor. Therefore, the current flowing through the \( 0.3 \Omega \) resistor is \( 4 \, A \). This is justified because in a series circuit, the current is constant throughout all components.