Question:

Consider the circuit shown below. A 12V battery is connected to a resistor with resistance R, and the resistor is in series with an unknown component X. The current passing through the circuit is measured as 2A.

1. Determine the equivalent resistance of the circuit.
2. Calculate the power dissipated by the resistor.
3. Determine the values of resistance R and component X.

Answer:

1. To determine the equivalent resistance of the circuit, we need to consider that resistors in series add up. Since we only have one resistor in series with the component X, the equivalent resistance can be calculated as follows:

   Req = R + RX
   

2. The power dissipated by a resistor can be calculated using the formula:

   P = I^2 * R
   

   where I is the current passing through the resistor and R is the resistance. In this case, the current passing through the resistor R is 2A. So, the power dissipated by the resistor can be calculated as:

   P = (2A)^2 * R
   

3. To determine the values of resistance R and component X, we need to use the equations derived from Kirchhoff's laws. First, we will calculate the resistance values.

   Using Kirchhoff's voltage law (KVL), the sum of the voltage drops across the resistor R and component X should be equal to the battery voltage:

   VR + VX = V
   R * I + RX * I = V
   

   Substituting the given values V = 12V and I = 2A, we get:

   2R + 2RX = 12
   

   Next, let's combine this equation with the equivalent resistance equation:

   Req = R + RX
   

   We can now substitute the value of Req into the previous equation:

   2(Req - RX) + 2RX = 12
   2Req = 12
   Req = 6Ω
   

   Now we have an equation with only one variable RX:

   6 = RX
   RX = 6Ω
   

   Therefore, the resistance R is 0Ω and the resistance of component X is 6Ω.

   • Resistance R: 0Ω
   • Resistance of component X: 6Ω

   It is important to note that a resistor with zero resistance represents a perfect conductor, meaning it has negligible resistance.