Question

Two resistors of 15 and 30 ω are connected in parallel. if the combination is connected in series with a 9. 0-v battery and a 20-ω resistor, what is the current through the 15-ω resistor?

Answer 1

The current through the 15 Ω resistor is 0.3 A.

Step-by-step explanation:

In this circuit, the 15 Ω and 30 Ω resistors are connected in parallel. So, we can use the formula for total resistance in parallel which is:

Req = (R1 * R2) / (R1 + R2)

Substituting the values, we get:

Req = (15 Ω * 30 Ω) / (15 Ω + 30 Ω) = 10 Ω

Now, the combination of the two resistors is connected in series with a 9.0 V battery and a 20 Ω resistor. The total resistance in series is the sum of all resistances:

Rtotal = Req + R3

Substituting the values, we get:

Rtotal = 10 Ω + 20 Ω = 30 Ω

Using Ohm's Law, we can calculate the current through the circuit:

I = V / R

Substituting the values, we get:

I = 9.0 V / 30 Ω = 0.3 A

Since the resistors are connected in parallel, the current through each resistor will be the same. Therefore, the current through the 15 Ω resistor is also 0.3 A.

Answer 2

To find the current through the 15-Ω resistor, we first calculate the equivalent resistance of the parallel combination (10 Ω) and then the total circuit resistance (30 Ω). Using Ohm's law, the total current is determined (0.3 A), and since the voltage across the parallel resistors is the same, the current through the 15-Ω resistor is 0.6 A.

Step-by-step explanation:

The problem involves a circuit with two resistors in parallel, one having a resistance of 15 Ω and the other 30 Ω. This parallel combination is then connected in series with a 20-Ω resistor and powered by a 9.0-V battery. To find the current through the 15-Ω resistor, we first need to calculate the equivalent resistance of the parallel combination.

The equivalent resistance (Req,parallel) of resistors in parallel can be found using the formula:
1 / Req,parallel = 1 / R1 + 1 / R2. In this case:
1 / R_eq,parallel = 1 / 15 Ω + 1 / 30 Ω = 1 / 10 Ω. Hence, R_eq,parallel = 10 Ω.

Now, the total equivalent resistance of the entire circuit, which is R_total = R_eq,parallel + R_series, becomes 10 Ω + 20 Ω = 30 Ω.

Using Ohm's law, the total current (I_total) from the battery can be found by I_total = V / R_total = 9.0 V / 30 Ω = 0.3 A. Since the resistors are in parallel, the voltage across them is the same, and it's equal to the voltage of the battery. So, the current through the 15-Ω resistor (I15Ω) would also be found using Ohm's law: I15Ω = V / R15Ω = 9.0 V / 15 Ω = 0.6 A. Therefore, the current through the 15-Ω resistor is 0.6 A.