Electric Circuits Exam Question

A circuit consists of a 12V battery, two resistors (R1 = 4Ω and R2 = 6Ω) connected in series, and a switch. The switch is initially closed, and a current flows through the circuit. When the switch is opened, find:

a) The initial current flowing through the circuit. b) The total resistance of the circuit. c) The final current flowing through the circuit after the switch is opened.

Answer:

a) The initial current flowing through the circuit: We know that the current flowing through a series circuit is the same at any point. Therefore, the initial current flowing through the circuit can be found using Ohm's Law:

$I = \frac{V}{R_{\text{total}}}$

where: $V = 12\, V$ (battery voltage) $R_{\text{total}} = R1 + R2 = 4\, \Omega + 6\, \Omega = 10\, \Omega$ (total resistance of the series circuit)

$I = \frac{12\, V}{10\, \Omega} = 1.2\, A$

Therefore, the initial current flowing through the circuit is 1.2 A.

b) The total resistance of the circuit: The total resistance of a series circuit is the sum of the individual resistances. Therefore, the total resistance of this circuit is:

$R_{\text{total}} = R1 + R2 = 4\, \Omega + 6\, \Omega = 10\, \Omega$

Therefore, the total resistance of the circuit is 10 Ω.

c) The final current flowing through the circuit after the switch is opened: When the switch is opened, the circuit becomes an open circuit, meaning no current can flow. Therefore, the final current flowing through the circuit after the switch is opened is 0 A.

In summary: a) The initial current flowing through the circuit is 1.2 A. b) The total resistance of the circuit is 10 Ω. c) The final current flowing through the circuit after the switch is opened is 0 A.