Question

In an electric circuit, three resistors \( 5 \, \Omega \), \( 10 \, \Omega \) and \( 15 \, \Omega \) are connected in series across a \( 60 \, V \) battery. Then the current flowing in the circuit is:

• A. \( 0.5 \, A \)
• B. \( 2 \, A \)
• C. \( 90 \, A \)
• D. \( 30 \, A \)

Hint:

Key concepts for series circuits: - The current is the same through all components. - The total resistance is the sum of individual resistances. - The total voltage across the circuit is the sum of the voltages across each component.

Answer 1

The Correct Option is B

Step 1: Determine the equivalent resistance of the series circuit.
When resistors are connected in series, their equivalent resistance ($R_{eq}$) is the sum of their individual resistances: \[ R_{eq} = R_1 + R_2 + R_3 \] Given \( R_1 = 5 \, \Omega \), \( R_2 = 10 \, \Omega \), and \( R_3 = 15 \, \Omega \), \[ R_{eq} = 5 \, \Omega + 10 \, \Omega + 15 \, \Omega = 30 \, \Omega \] Step 2: Apply Ohm's Law to find the current.
Ohm's Law states that the current ($I$) flowing through a conductor between two points is directly proportional to the voltage ($V$) across the two points and inversely proportional to the resistance ($R$) of the conductor: \[ V = I \times R \quad \Rightarrow \quad I = \frac{V}{R} \] In this circuit, the voltage across the equivalent resistance is the voltage of the battery, \( V = 60 \, V \), and the equivalent resistance is \( R_{eq} = 30 \, \Omega \). Step 3: Calculate the current \( I \). \[ I = \frac{V}{R_{eq}} = \frac{60 \, V}{30 \, \Omega} = 2 \, A \] The current flowing in the circuit is \( 2 \, A \).