A DC circuit to a specialized of 3000 ohms, a 500 ohms, and 10 000 ohms resistor in series. The series current is 15 mA. Find the Voltage drop across R1.
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R1 = 3,000 Ohms = 3k Ohms
R2 = 0.5k
R3 = 10k
I = 15 mA
V1 = I*R1 = 15mA * 3k = 45 V.
To find the voltage drop across R1, we need to use Ohm's Law, which states that the voltage drop (V) across a resistor is equal to the current (I) flowing through it multiplied by the resistance (R).
First, let's determine the total resistance in the circuit, which is the sum of the resistances in series:
Total resistance (Rt) = R1 + R2 + R3
= 3000 ohms + 500 ohms + 10000 ohms
= 13300 ohms
Next, we'll use Ohm's Law to find the voltage drop across R1:
V1 = I * R1
Given:
Current (I) = 15 mA = 0.015 A (Convert mA to A)
Resistance (R1) = 3000 ohms
V1 = 0.015 A * 3000 ohms
= 45 V
Therefore, the voltage drop across R1 is 45 volts.