Question

A circuit containing several resistors is connected to a 9.0 V battery. The circuit has 4 mA of current passing through it.

To the nearest hundredth of a kΩ, what is the resistance

Answer 1

To calculate the resistance R1, we use Ohm's Law with the information given for the voltages and currents when different batteries are added in series. By creating equations from Ohm's Law and solving for R1, we find that the resistance is 90 ohms.

Step-by-step explanation:

To find the resistance of R1, we will use Ohm's Law, which states that the voltage (V) across a resistor is equal to the current (I) through the resistor times the resistance (R). Mathematically, this is expressed as V = IR.

When the second battery was added, the total voltage became V1 + 9.0 V, and the current measured was 0.3 A. Using Ohm's Law, we have:
(1) V1 + 9.0 V = 0.3 A * R1

When the third battery was added, the total voltage became V1 + 9.0 V + 9.0 V, with a current of 0.4 A. Using Ohm's Law again, we have:
(2) V1 + 18.0 V = 0.4 A * R1

Subtracting equation (1) from equation (2) eliminates V1, allowing us to solve for R1:

(2) - (1) => (V1 + 18.0 V) - (V1 + 9.0 V) = (0.4 A * R1) - (0.3 A * R1)

9.0 V = 0.1 A * R1 => R1 = 9.0 V / 0.1 A => R1 = 90 Ω

So, the resistance of R1 is 90 ohms.