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9. A 310 2, a 500 , and a 34022 resistor are connected in series across a 12V battery. What is the

total resistance of the circuit? What is the current through each resistor? What us the voltage
drop across each resistor?

User Ingyesid
by
8.1k points

1 Answer

3 votes

Answer:

Total Resistance ≈ 34832 Ω

Total Current ≈ 0.000344 A

Voltage Drop across R1 ≈ 0.10624 V

Voltage Drop across R2 ≈ 0.172 V

Voltage Drop across R3 ≈ 11.70448 V

Step-by-step explanation:

To find the total resistance of a series circuit, you simply add up the individual resistances:

Total Resistance (R_total) = R1 + R2 + R3

Given the resistances:

R1 = 310 Ω

R2 = 500 Ω

R3 = 34022 Ω

Total Resistance = 310 Ω + 500 Ω + 34022 Ω

Total Resistance ≈ 34832 Ω

Now, to find the current through each resistor in a series circuit, you can use Ohm's Law:

Current (I) = Voltage (V) / Resistance (R)

Since the battery voltage is 12V and the total resistance is 34832 Ω, you can calculate the total current:

Total Current = 12 V / 34832 Ω

Total Current ≈ 0.000344 A

Since the circuit is a series circuit, the same current flows through all the resistors:

Current through each resistor = Total Current

Current through each resistor ≈ 0.000344 A

To find the voltage drop across each resistor, you can use Ohm's Law again:

Voltage Drop (V_drop) = Current (I) × Resistance (R)

For each resistor:

Voltage Drop across R1 = Current × R1

Voltage Drop across R2 = Current × R2

Voltage Drop across R3 = Current × R3

Voltage Drop across R1 ≈ 0.000344 A × 310 Ω

Voltage Drop across R2 ≈ 0.000344 A × 500 Ω

Voltage Drop across R3 ≈ 0.000344 A × 34022 Ω

Voltage Drop across R1 ≈ 0.10624 V

Voltage Drop across R2 ≈ 0.172 V

Voltage Drop across R3 ≈ 11.70448 V

User RenanStr
by
8.6k points